7
Formal proofs are a convenient and powerful way to test arguments. They also help develop our reasoning skills. From now on, formal proofs will be our main method of testing arguments.
7.1 Easier proofs
A formal proof breaks an argument into a series of small steps. We’ll use an indirect proof strategy, whereby we first assume the opposite of what we want to prove. You may remember such proofs from high-school geometry; to prove that two angles are equal, assume that they aren’t equal – and then show that this is impossible, because it leads to a contradiction. Similarly, to prove that the butler committed the murder, assume that he didn’t do it – and then show that this is impossible, because it leads to a contradiction.
Here’s an English analog of a formal proof. Suppose we know premises 1 to 4 and want to prove from them that the butler committed the murder:
· 1 The only people in the mansion were the butler and the maid.
· 2 If the only people in the mansion were the butler and the maid, then the butler or the maid did it.
· 3 If the maid did it, then she had a motive.
· 4 The maid didn’t have a motive.
· ∴ The butler did it.
· 1 T
· 2 (T ⊃ (B ∨ M))
· 3 (M ⊃ H)
· 4 ∼H
· ∴ B
First assume that the butler didn’t do it (∼B). From 1 and 2, conclude that the butler or the maid did it (B ∨ M). From 3 and 4, conclude that the maid didn’t do it (∼M). From these last two, conclude that the butler did it (B). This contradicts our assumption, that the butler didn’t do it, which is then shown to be false; so therefore, given premises 1 to 4, the butler did it. So the butler is guilty – throw him in jail!
A formal proof is like this, but in symbols. For now, we’ll use a three-step strategy: (1) START (assume the conclusion’s opposite), (2) S&I (derive further lines using S- and I-rules until we get a contradiction), and (3) RAA (derive the original conclusion). We START this way with our butler argument: 0147
· 1 T
· 2 (T ⊃ (B ∨ M))
· 3 (M ⊃ H)
· 4 ∼H
· [ ∴ B
· 5 asm: ∼B
In the START step here, we block off the conclusion “B” (which reminds us not to use it in deriving further lines) and add “asm:” (for “assume”) followed by its simpler contradictory, “∼B.”
We begin the S&I step by glancing at the complex wffs (any wffs longer than a single letter or its negation) and noticing their forms; here the complex wffs are 2 and 3, both IF-THENs. Recall that AND, NOR, and NIF simplify using S-rules, while NOT-BOTH, OR, and IF-THEN can infer using I-rules, if certain extra wffs are available. Since the complex lines here, 2 and 3, are IF-THENs, we can infer with each if we have the first part true or the second part false – which we do have (2 has 1, and 3 has 4). So we derive further formulas:
· 1 T
· * 2 (T ⊃ (B ∨ M))
· * 3 (M ⊃ H)
· 4 ∼H
· [ ∴ B
· 5 asm: ∼B
· * 6 ∴ (B ∨ M) {from 1 and 2}
· 7 ∴ M {from 5 and 6}
· 8 ∴ H {from 3 and 7}
Lines 1 and 2 give us line 6 by an IF-THEN rule: “IF-THEN, affirm the first, so affirm the second.” Likewise, lines 5 and 6 give us line 7 by the OR rule: “At least one is true, this one isn’t, so the other one is.” Finally, lines 3 and 7 give us line 8 by an IF-THEN rule: “IF-THEN, affirm the first, so affirm the second.” And so we get a contradiction between line 4 and 8 (“∼H” and “H”).
Here we starred lines 2, 3, and 6 when we used them to derive further formulas. Starring a line tells us that we’ve used it and so it can be somewhat ignored as we try to derive further steps. I’ll talk more about this later.
Once we get a contradiction, as between lines 4 and 8 above, we finish the proof using RAA (reductio ad absurdum, reduction to absurdity), which roughly says that an assumption that leads to a contradiction is thereby wrong, and so we can conclude the opposite – which is our original conclusion. At the same time, we block off the lines from the last assumption on down to show that they can’t be used in deriving further lines. This finishes our first formal proof: 0148
Now that we’ve seen a complete proof, we need to firm up the details.
(1) START: Start a proof by blocking off the original conclusion (blocking off tells us to ignore a line for the rest of the proof) and assuming its simpler contradictory. Two wffs are contradictories if they are exactly alike except that one starts with an additional “∼.” So if our conclusion is “A,” then assume “∼A”; but if our conclusion is “∼A,” then assume “A.” And if our conclusion is “(A ⊃ B),” then assume “∼(A ⊃ B).” Always add or subtract an initial squiggle to the original conclusion.
(2) S&I: Derive further lines using S- and I-rules until there’s a contradiction. Focus on complex wffs that aren’t starred or blocked off. Note the forms of these wffs: AND, NOR, and NIF can be simplified, while NOT-BOTH, OR, and IF-THEN can infer if certain other wffs are available. In our sample proof, our first inference has to involve lines 2 or 3, both IF-THENs and the only complex wffs. Often we can do a proof in various ways; so instead of deriving “H” in line 8, we could use 3 and 4 to get “∼M,” which would contradict 7.
We starred lines 2, 3, and 6. Here are the starring rules – with examples:
Star any wff simplified using an S-rule.
* (A • B)
––––––––
∴ A
∴ B
Star the longer wff used in an I-rule inference.
* (A ⊃ B)
A
––––––––
∴ B
Starred lines are redundant, since shorter lines have the same information. When you do a proof, focus on complex wffs that aren’t starred or blocked off and what can be derived from them.1 While starring is optional, it simplifies your work because it leads you to ignore lines that won’t help to derive further formulas. 0149
1 Once you’ve starred a complex wff, it’s pointless to again use an S-rule on it or to again use it as the longer wff in an I-rule inference. So we can focus on complex wffs that aren’t starred or blocked off. But it may be useful to use a starred wff as the smaller wff in an I-rule inference. Suppose you starred “(A • B)” when you simplified it. If you have an unstarred “((A • B) ⊃ C),” feel free to combine it with the starred “(A • B)” to derive “C.”
In the S&I part, we’ll use these old S- and I-rules (these and the three new rules hold regardless of what pairs of contradictory wffs replace “P” / “∼P” and “Q” / “∼Q”):
|
AND |
(P • Q) –––––– P, Q |
|
NOR |
~(P ∨ Q) –––––– ~P, ~Q |
|
NIF |
~(P ⊃ Q) –––––– P, ~Q |
|
NOT-BOTH |
|
|
~(P • Q) P –––––– ~Q |
~(P • Q) Q –––––– ~P |
|
OR |
|
|
(P ∨ Q) ~P –––––– Q |
(P ∨ Q) ~Q –––––– P |
|
IF-THEN |
|
|
(P ⊃ Q) P –––––– Q |
(P ⊃ Q) ~Q –––––– ~P |
And we’ll add three new S-rules – NN, IFF, and NIFF – which we won’t use much:
NN
∼∼P
––––
P
NN (NOT-NOT, double negation) eliminates “∼∼” from the beginning of a wff.
IFF
(P ≡ Q)
––––––––––––––
(P ⊃ Q), (Q ⊃ P)
IFF breaks a biconditional into two conditionals.
NIFF
∼(P ≡ Q)
––––––––––––––
(P ∨ Q), ∼(P • Q)
NIFF1 breaks up the denial of a biconditional; since “(P ≡ Q)” says that P and Q have the same truth value, “∼(P ≡ Q)” says that P and Q have different truth values – so one or the other is true, but not both.2
1 To avoid confusion, pronounce “NIFF” as “knife” and “IFF” with a long “i” to rhyme with this.
2 The S-rules also work in the other direction (so “(A • B)” follows from “A” and “B”); but our proofs standardly use S-rules only to simplify. The LogiCola software lets you use two further rules: (1) Given “(A ≡ B)”: if you have one side true, you can get the other true – and if you have one side false, you can get the other false. (2) Given “∼(A ≡ B)”: if you have one side true, you can get the other false – and if you have one side false, you can get the other true.
In applying S- and I-rules, look for lines of these forms to simplify:
AND NOR NIF
NN IFF NIFF
or a pair of lines to infer from:
NOT-BOTH (with one part true)
OR (with one part false)
IF-THEN (with part-1 true or part-2 false)
Note that there’s a rule for each of the nine possible complex wff forms.
Here’s another summary of the S- and I-rules (here “→” means we can infer whole lines from left to right): 0150
S-rules (Simplifying)
AND, NOR, NIF, NN, IFF, NIFF
(P • Q) → P, Q
∼(P ∨ Q) → ∼P, ∼Q
∼(P ⊃ Q) → P, ∼Q
∼∼P → P
(P ≡ Q) → (P ⊃ Q), (Q ⊃ P)
∼(P ≡ Q) → (P ∨ Q), ∼(P • Q)
I-rules (Inferring)
NOT-BOTH, OR, IF-THEN
∼(P • Q), P → ∼Q
∼(P • Q), Q → ∼P
(P ∨ Q), ∼P → Q
(P ∨ Q), ∼Q → P
(P ⊃ Q), P → Q
(P ⊃ Q), ∼Q → ∼P
Read “(P • Q) → P, Q” as “from ‘(P • Q)’ one may derive ‘P’ and also ‘Q.’” As you learn formal proofs, it’s good to practice the S- and I-rules.
(3) RAA: Rule RAA says roughly that an assumption is false if it leads to contradictory wffs (a pair, like “H” and “∼H,” that’s identical except that one starts with an additional squiggle). The contradictory wffs may occur anywhere in the proof (as premises, assumptions, or derived lines), as long as neither is blocked off. Here’s a more precise formulation of RAA:
RAA: Suppose some pair of not-blocked-off lines has contradictory wffs. Then block off all the lines from the last not-blocked-off assumption on down and infer a line consisting in “∴” followed by a contradictory of that assumption.
Blocking off forbids deriving further lines using the assumption (which now is shown to be false). This is important later, with multiple-assumption proofs.
Here are some key definitions about formal proofs:
· A premise is a line consisting of a wff by itself (with no “asm:” or “∴”).
· An assumption is a line consisting of “asm:” and then a wff.
· A derived line is a line consisting of “∴” and then a wff.
· A formal proof is a vertical sequence of zero or more premises followed by one or more assumptions or derived lines, where each derived line follows from previously not-blocked-off lines by one of the S- and I-rules listed above or by RAA, and each assumption is blocked off using RAA.
By the last definition, the stars, line numbers, blocked off original conclusion, and justifications aren’t strictly part of the proof. Instead, these are unofficial helps – and some people skip them. On the other hand, some people like to mention the inference rule (like “AND” or “IF-THEN”) in the justifications; so then a justification might say “{from 1 and 2 using IF-THEN}.” If you’re taking a logic course, follow your teacher’s directives about such matters.
A wff is a theorem if it’s provable from zero premises. Here’s a premiseless proof (it’s valid because the conclusion is a logically necessary truth): 0151
Again, our proof strategy has three steps. (1) START: Block off the conclusion and assume its contradictory (line 1). (2) S&I: Derive lines 2 to 4 and get a contradiction. (3) RAA: Use RAA to finish the proof (line 5). Our proof strategy gets more complex later, with invalid arguments and multiple assumptions.
A formal proof, as we defined it, must use the specified S- and I-rules or RAA to derive further lines. We can’t just use any intuitive inferences that we think will work (although advanced users sometimes take such shortcuts1). There can be legitimate variations in how to do proofs. So one person might always simplify “(A • B)” into the two parts, “A” first and then “B.” Another might derive “B” first and then “A.” Yet another person might derive just the part needed to get a contradiction. All three approaches are fine and allowed by the LogiCola computer proof exercises.
1 An example of a shortcut is to infer “C” immediately from previous lines “(A ⊃ (B ⊃ C))” and “A” and “B” in a single step. Don’t take such shortcuts unless your teacher allows them and you’re very sure that your formula validly follows, even though not licensed by our rules. I suggest that you thoroughly master our normal strategy before taking shortcuts; otherwise, your reliance on shortcuts may lead to steps that don’t validly follow and may prevent you from learning our proof procedure (which will always work if you do it right). LogiCola doesn’t allow shortcuts.
Why not add further inference rules to our system, since this would shorten some proofs?
The downside is that this would also make our system harder to learn. Our proof system was designed in a practical way, to produce reasonably short proofs and yet be easy to learn and use.
LogiCola proofs begin by giving you a randomly generated problem (there are many millions of possible problems). You keep giving the next line until the problem is done. You can vary the kind of problem: Easier / Harder / Mixed – and Valid / Invalid / Combined. As you begin, turn on training wheels; this gives you suggestions about what to do next – but these suggestions disappear as you make progress in the exercise. You can also have the program automatically star lines that you’ve used, or you can choose to star yourself (but you don’t lose points for getting these wrong). You can click (or touch) an arrow at the top to give you the next line or to finish the problem (but without getting credit for the problem); some students use these arrows to step through sample proofs before starting them on their own. You can click (or touch) a previous line to copy it into the answer space, so you can then modify it to give your next line. If you’re new to proofs, I suggest you read the LogiCola help-section on proofs.
7.1a Exercise: LogiCola F (TE & TH) and GEV
Prove each of these arguments to be valid (all are valid). 0152
· 1. (A ⊃ B)
∴ (∼B ⊃ ∼A)
· 2. A
∴ (A ∨ B)
· 3. (A ⊃ B)
(∼A ⊃ B)
∴ B
· 4. ((A ∨ B) ⊃ C)
∴ (∼C ⊃ ∼B)
· 5. (A ∨ B)
(A ⊃ C)
(B ⊃ D)
∴ (C ∨ D)
· 6. (A ⊃ B)
(B ⊃ C)
∴ (A ⊃ C)
· 7. (A ≡ B)
∴ (A ⊃ (A • B))
· 8. ∼(A ∨ B)
(C ∨ B)
∼(D • C)
∴ ∼D
· 9. (A ⊃ B)
∼B
∴ (A ≡ B)
· 10. (A ⊃ (B ⊃ C))
∴ ((A • B) ⊃ C)
7.1b Exercise: LogiCola F (TE & TH) and GEV
First appraise intuitively. Then translate into logic (using the letters given) and prove to be valid (all are valid).
1. If Heather saw the butler putting the tablet into the drink and the tablet was poison, then the butler killed the deceased.
Heather saw the butler putting the tablet into the drink.
∴ If the tablet was poison, then the butler killed the deceased. [Use H, T, and B.]
2. If we had an absolute proof of God’s existence, then our will would be irresistibly attracted to do right.
If our will were irresistibly attracted to do right, then we’d have no free will.
∴ If we have free will, then we have no absolute proof of God’s existence. [Use P, I, and F; from Immanuel Kant and John Hick, who used it to explain why God doesn’t make his existence more evident.]
3. If racism is clearly wrong, then either it’s factually clear that all races have equal abilities or it’s morally clear that similar interests of all beings ought to be given equal consideration.
It’s not factually clear that all races have equal abilities.
If it’s morally clear that similar interests of all beings ought to be given equal consideration, then similar interests of animals and humans ought to be given equal consideration.
∴ If racism is clearly wrong, then similar interests of animals and humans ought to be given equal consideration. [Use W, F, M, and A. This argument is from Peter Singer, who fathered the animal liberation movement.] 0153
4. The universe is orderly (like a watch that follows complex laws).
Most orderly things we’ve examined have intelligent designers.
We’ve examined a large and varied group of orderly things.
If most orderly things we’ve examined have intelligent designers and we’ve examined a large and varied group of orderly things, then probably most orderly things have intelligent designers.
If the universe is orderly and probably most orderly things have intelligent designers, then the universe probably has an intelligent designer.
∴ The universe probably has an intelligent designer. [Use U, M, W, P, and D. This is a form of the argument from design for the existence of God.]
5. If God doesn’t want to prevent evil, then he isn’t all good.
If God isn’t able to prevent evil, then he isn’t all powerful.
Either God doesn’t want to prevent evil, or he isn’t able.
∴ Either God isn’t all powerful, or he isn’t all good. [Use W, G, A, and P. This form of the problem-of-evil argument is from the ancient Greek Empiricus.]
6. If Genesis gives the literal facts, then birds were created before humans. (Genesis 1:20–26)
If Genesis gives the literal facts, then birds weren’t created before humans. (2:5–20)
∴ Genesis doesn’t give the literal facts. [Use L and B. Origen, an early Christian thinker, gave similar textual arguments against taking Genesis literally.]
7. The world had a beginning in time.
If the world had a beginning in time, there was a cause for the world’s beginning.
If there was a cause for the world’s beginning, a personal being caused the world.
∴ A personal being caused the world. [Use B, C, and P. This “Kalam argument” for the existence of God is from William Craig and James Moreland; they defend premise 1 by various considerations, including the Big Bang theory, the law of entropy, and the impossibility of an actual infinite.]
8. If the world had a beginning in time and it didn’t just pop into existence without any cause, then the world was caused by God.
If the world was caused by God, then there is a God.
There is no God.
∴ Either the world had no beginning in time, or it just popped into existence without any cause. [Use B, P, C, and G; from J. L. Mackie, who based his “There is no God” premise on the problem-of-evil argument.]
9. Closed systems tend toward greater entropy (a more randomly uniform distribution of energy). (This is the second law of thermodynamics.)
If closed systems tend toward greater entropy and the world has existed through endless time, then the world would have achieved almost complete entropy (for example, everything would be about the same temperature).
The world has not achieved almost complete entropy.
If the world hasn’t existed through endless time, then the world had a beginning in time.
∴ The world had a beginning in time. [Use G, E, C, and B; from William Craig and James Moreland.] 0154
10.If time stretches back infinitely, then today wouldn’t have been reached.
If today wouldn’t have been reached, then today wouldn’t exist.
Today exists.
If time doesn’t stretch back infinitely, then there was a first moment of time.
∴ There was a first moment of time. [I, R, T, F]
11.If there are already laws preventing discrimination against women, then if the Equal Rights Amendment (ERA) would rob women of many current privileges then it is the case both that passage of the ERA would be against women’s interests and that women ought to work for its defeat.
The ERA would rob women of many current privileges (like draft exemption).
∴ If there are already laws preventing discrimination against women, then women ought to work for the defeat of the ERA. [L, R, A, W]
12.If women ought never to be discriminated against, then we should pass current laws against discrimination and block future discriminatory laws against women.
The only way to block future discriminatory laws against women is to pass an Equal Rights Amendment (ERA).
If we should block future discriminatory laws against women and the only way to do this is to pass an ERA, then we ought to pass an ERA.
∴ If women ought never to be discriminated against, then we ought to pass an ERA. [N, C, F, O, E]
13.If the claim that knowledge-is-impossible is true, then we understand the word “know” but there are no cases of knowledge.
If we understand the word “know,” then the meaning of “know” comes either from a verbal definition or from experienced examples of knowledge.
If the meaning of “know” comes from a verbal definition, then there’s an agreed-upon definition of “know.”
There’s no agreed-upon definition of “know.”
If the meaning of “know” comes from experienced examples of knowledge, then there are cases of knowledge.
∴ The claim that knowledge-is-impossible is false. [Use I, U, C, D, E, and A. This is a form of the paradigm-case argument.]
14.If p is the greatest prime, then n (we may stipulate) is one plus the product of all the primes less than p.
If n is one plus the product of all the primes less than p, then either n is prime or else n isn’t prime but has prime factors greater than p.
If n is prime, then p isn’t the greatest prime.
If n has prime factors greater than p, then p isn’t the greatest prime.
∴ p isn’t the greatest prime. [Use G, N, P, and F. This proof that there’s no greatest prime number is from the ancient Greek mathematician Euclid.]
7.2 Easier refutations
This example shows how our proof strategy works with an invalid argument: 0155
· The only people in the mansion were the butler and the maid.
· If the only people in the mansion were the butler and the maid, then the butler or the maid did it.
· If the maid did it, then she had a motive.
· ∴ The butler did it.
· T
· (T ⊃ (B ∨ M))
· (M ⊃ H)
· ∴ B
The butler’s lawyer could object: “Yes, the only people in the mansion were the butler and the maid, and so one of them did the killing. But maybe the maid had a motive and did it, instead of the butler. The known facts are consistent with this possibility and so don’t show that the butler did it.” This is a refutation – a set of possible truth conditions making the premises all true and conclusion false. A refutation shows that the argument is invalid.
If we try to prove this invalid argument, we’ll assume the conclusion’s opposite and then use S- and I-rules to derive whatever we can:
· 1 T
· * 2 (T ⊃ (B ∨ M))
· * 3 (M ⊃ H)
· [ ∴ B
· 4 asm: ∼B
· * 5 ∴ (B ∨ M) {from 1 and 2}
· 6 ∴ M {from 4 and 5}
· 7 ∴ H {from 3 and 6}
We can derive no contradiction. So we instead construct a refutation box – which contains the simple wffs (letters or their negations) from not-blocked-off lines (1, 4, 6, and 7) – and we plug its truth values into the original argument:
· 1 T1 = 1 Invalid
· 2 (T1 ⊃ (B0 ∨ M1)) = 1
· 3 (M1 ⊃ H1) = 1
· [ ∴ B0 = 0
T, M, H, ∼B
These truth conditions make the premises all true and conclusion false. This shows that the argument is invalid.
With invalid arguments, we don’t get a contradiction; instead, we get a refutation. To construct the refutation box, take the simple wffs (letters or their negation) from not-blocked-off lines and put them in a box (their order doesn’t matter). Our box also could be written in either of these two ways:
T = 1, M = 1, H = 1, B = 0
T1, M1, H1, B0
Then plug the truth values into the original argument. If the refutation box has a letter by itself (like “T” or “M”), then mark that letter true (“1”) in the 0156 argument; if it has the negation of a letter (like “∼B”), then mark that letter false (“0”); any letters that don’t occur in the box are unknown (“?” – the refutation may still work). Then see if these values make the premises all true and conclusion false; if they do, then that shows that the argument is invalid.
If we don’t get the premises all true and conclusion false, then we did something wrong. The faulty line (a premise that’s false or unknown, or a conclusion that’s true or unknown) is the problem’s source; maybe we derived something from it wrongly, or didn’t derive something we should have derived. So our strategy tells us if something goes wrong and where to look to fix the problem.
Let me summarize. Suppose we want to show that, given certain premises, the butler must be guilty. We assume that he’s innocent and try to show that this leads to a contradiction. If we get a contradiction, then his innocence is impossible and so he must be guilty. But if we get no contradiction, then we may be able to show how the premises could be true while yet he is innocent, thus showing that the argument against him is invalid.
Here’s another invalid argument and its refutation:
· 1 (A0 ⊃ B1) = 1 Invalid
· * 2 (C0 ∨ B1) = 1
· [ ∴ (C0 ∨ A0) = 0
· * 3 asm: ∼(C ∨ A)
· 4 ∴ ∼C {from 3}
· 5 ∴ ∼A {from 3}
· 6 ∴ B {from 2 and 4}
B, ∼A, ∼C
We get nothing from “(A ⊃ B)” in line 1, since we’d need “A” true or “B” false. So we’ve derived all we can. Since we have no contradiction, we construct a refutation box. We plug the values into the argument and get the premises all true and conclusion false. This shows that the argument is invalid.
Our proof strategy so far looks like this (we’ll add another step later):
1. START: Block off the conclusion and add “asm:” followed by the conclusion’s simpler contradictory.
2. S&I: Go through the complex wffs that aren’t starred or blocked off and use these to derive new wffs using S- and I-rules. Star any wff you simplify using an S-rule, or the longer wff used in an I-rule inference. If you get a contradiction, then go to RAA (step 3). If you can’t derive anything further and yet have no contradiction, then go to REFUTE (step 4).
3. RAA: Apply the RAA rule. You’ve proved the argument valid.
4. REFUTE: Construct a refutation box containing any simple wffs (letters or their negation) that aren’t blocked off. In the original argument, mark each letter “1” or “0” or “?” depending on whether the box has the letter or its negation or neither. If these truth conditions make the premises all true and conclusion false, then this shows the argument to be invalid. 0157
7.2a Exercise: LogiCola GEI
Prove each of these arguments to be invalid (all are invalid).
(A ⊃ B)
∴ (B ⊃ A)
· 1 (A0 ⊃ B1) = 1 Invalid
· [ ∴ (B1 ⊃ A0) = 0
· * 2 asm: ∼(B ⊃ A)
· 3 ∴ B {from 2}
· 4 ∴ ∼A {from 2}
B, ∼A
1. (A ∨ B)
∴ A
2. (A ⊃ B)
(C ⊃ B)
∴ (A ⊃ C)
3. ∼(A • ∼B)
∴ ∼(B • ∼A)
4. (A ⊃ (B • C))
(∼C ⊃ D)
∴ ((B • ∼D) ⊃ A)
5. ((A ⊃ B) ⊃ (C ⊃ D))
(B ⊃ D)
(A ⊃ C)
∴ (A ⊃ D)
6. (A ≡ B)
(C ⊃ B)
∼(C • D)
D
∴ ∼A
7. ((A • B) ⊃ C)
∴ (B ⊃ C)
8. ((A • B) ⊃ C)
((C ∨ D) ⊃ ∼E)
∴ ∼(A • E)
9. ∼(A • B)
(∼A ∨ C)
∴ ∼(C • B)
10.∼(∼A • ∼B)
∼C
(D ∨ ∼A)
((C • ∼E) ⊃ ∼B)
∼D
∴ ∼E
7.2b Exercise: LogiCola GEC
First appraise intuitively. Then translate into logic (using the letters given) and say whether valid (and give a proof) or invalid (and give a refutation).
1. If the butler shot Jones, then he knew how to use a gun.
If the butler was a former marine, then he knew how to use a gun.
The butler was a former marine.
∴ The butler shot Jones. [Use S, K, and M.]
2. If virtue can be taught, then either there are professional virtue-teachers or there are amateur virtue-teachers.
If there are professional virtue-teachers, then the Sophists can teach their students to be virtuous.
If there are amateur virtue-teachers, then the noblest Athenians can teach their children to be virtuous.
The Sophists can’t teach their students to be virtuous and the noblest Athenians (such as the great leader Pericles) can’t teach their children to be virtuous.
∴ Virtue can’t be taught. [Use V, P, A, S, and N; from Plato’s Meno.] 0158
3. It would be equally wrong for a sadist (through drugs that would blind you but not hurt your mother) to have blinded you permanently before or after your birth.
If it would be equally wrong for a sadist (through such drugs) to have blinded you permanently before or after your birth, then it’s false that one’s moral right to equal consideration begins at birth.
If infanticide is wrong and abortion isn’t wrong, then one’s moral right to equal consideration begins at birth.
Infanticide is wrong.
∴ Abortion is wrong. [Use E, R, I, and A.]
4. If you hold a moral belief and don’t act on it, then you’re inconsistent.
If you’re inconsistent, then you’re doing wrong.
∴ If you hold a moral belief and act on it, then you aren’t doing wrong. [Use M, A, I, and W. Is the conclusion plausible? What more plausible conclusion follows from these premises?]
5. If Socrates escapes from jail, then he’s willing to obey the state only when it pleases him.
If he’s willing to obey the state only when it pleases him, then he doesn’t really believe what he says and he’s inconsistent.
∴ If Socrates really believes what he says, then he won’t escape from jail. [Use E, W, R, and I; from Plato’s Crito. Socrates had been jailed and sentenced to death for teaching philosophy. He discussed with his friends whether he ought to escape from jail instead of suffering the death penalty.]
6. Either Socrates’s death will be perpetual sleep, or if the gods are good then his death will be an entry into a better life.
If Socrates’s death will be perpetual sleep, then he shouldn’t fear death.
If Socrates’s death will be an entry into a better life, then he shouldn’t fear death.
∴ Socrates shouldn’t fear death. [Use P, G, B, and F; from Plato’s Crito – except for which dropped premise?]
7. If predestination is true, then God causes us to sin.
If God causes us to sin and yet damns sinners to eternal punishment, then God isn’t good.
∴ If God is good, then either predestination isn’t true or else God doesn’t damn sinners to eternal punishment. [Use P, C, D, and G. This attacks the views of the American colonial thinker Jonathan Edwards.]
8. If determinism is true, then we have no free will.
If Heisenberg’s interpretation of quantum physics is correct, some events aren’t causally necessitated by prior events.
If some events aren’t causally necessitated by prior events, determinism is false.
∴ If Heisenberg’s interpretation of quantum physics is correct, then we have free will. [D, F, H, E] 0159
9. Government’s function is to protect life, liberty, and the pursuit of happiness.
The British colonial government doesn’t protect these.
The only way to change it is by revolution.
If government’s function is to protect life, liberty, and the pursuit of happiness and the British colonial government doesn’t protect these, then the British colonial government ought to be changed.
If the British colonial government ought to be changed and the only way to change it is by revolution, then we ought to have a revolution.
∴ We ought to have a revolution. [Use G, B, O, C, and R. This summarizes the reasoning behind the American Declaration of Independence. Premise 1 was claimed to be self-evident, premises 2 and 3 were backed by historical data, and premises 4 and 5 were implicit conceptual bridge premises.]
10.The apostles’ teaching either comes from God or is of human origin.
If it comes from God and we kill the apostles, then we will be fighting God.
If it’s of human origin, then it’ll collapse of its own accord.
If it’ll collapse of its own accord and we kill the apostles, then our killings will be unnecessary.
∴ If we kill the apostles, then either our killings will be unnecessary or we will be fighting God. [Use G, H, K, F, C, and U. This argument, from Rabbi Gamaliel in Acts 5:34–9, is perhaps the most complex reasoning in the Bible.]
11.If materialism (the view that only matter exists) is true, then idealism is false.
If idealism (the view that only minds exist) is true, then materialism is false.
If mental events exist, then materialism is false.
If materialists think their theory is true, then mental events exist.
∴ If materialists think their theory is true, then idealism is true. [M, I, E, T]
12.If determinism is true and cruelty is wrong, then the universe contains unavoidable wrong actions.
If the universe contains unavoidable wrong actions, then we ought to regret the universe as a whole.
If determinism is true and regretting cruelty is wrong, then the universe contains unavoidable wrong actions.
∴ If determinism is true, then either we ought to regret the universe as a whole (the pessimism option) or else cruelty isn’t wrong and regretting cruelty isn’t wrong (the “nothing matters” option). [Use D, C, U, O, and R. This sketches the reasoning in William James’s “The Dilemma of Determinism.” James thought that when we couldn’t prove one side or the other to be correct (as on the issue of determinism), it was more rational to pick our beliefs in accord with practical considerations. He argued that these weighed against determinism.]
13.If a belief is proved, then it’s worthy of acceptance.
If a belief isn’t disproved but is of practical value to our lives, then it’s worthy of acceptance.
If a belief is proved, then it’s not disproved.
∴ If a belief is proved or is of practical value to our lives, then it’s worthy of acceptance. [P, W, D, V] 0160
14.If you’re consistent and think that stealing is normally permissible, then you’ll consent to the idea of others stealing from you in normal circumstances.
You don’t consent to the idea of others stealing from you in normal circumstances.
∴ If you’re consistent, then you won’t think that stealing is normally permissible. [C, N, Y]
15.If the meaning of a term is always the object it refers to, then the meaning of “Fido” is Fido.
If the meaning of “Fido” is Fido, then if Fido is dead then the meaning of “Fido” is dead.
If the meaning of “Fido” is dead, then “Fido is dead” has no meaning.
“Fido is dead” has meaning.
∴ The meaning of a term isn’t always the object it refers to. [Use A, B, F, M, and H; from Ludwig Wittgenstein, except for which dropped premise?]
16.God is all powerful.
If God is all powerful, then he could have created the world in any logically possible way and the world has no necessity.
If the world has no necessity, then we can’t know the way the world is by abstract speculation apart from experience.
∴ We can’t know the way the world is by abstract speculation apart from experience. [Use A, C, N, and K; from the medieval William of Ockham.]
17.If God changes, then he changes for the worse or for the better.
If he’s perfect, then he doesn’t change for the worse.
If he changes for the better, then he isn’t perfect.
∴ If God is perfect, then he doesn’t change. [C, W, B, P]
18.If belief in God has scientific backing, then it’s rational.
No conceivable scientific experiment could decide whether there is a God.
If belief in God has scientific backing, then some conceivable scientific experiment could decide whether there is a God.
∴ Belief in God isn’t rational. [B, R, D]
19.Every event with finite probability eventually takes place.
If the nations of the world don’t get rid of their nuclear weapons, then there’s a finite probability that humanity will eventually destroy the world.
If every event with finite probability eventually takes place and there’s a finite probability that humanity will eventually destroy the world, then humanity will eventually destroy the world.
∴ Either nations of the world will get rid of their nuclear weapons, or humanity will eventually destroy the world. [E, R, F, H]
20.If the world isn’t ultimately absurd, then conscious life will go on forever and the world process will culminate in an eternal personal goal.
If there is no God, then conscious life won’t go on forever.
∴ If the world isn’t ultimately absurd, then there is a God. [Use A, F, C, and G; from the Jesuit scientist, Pierre Teilhard de Chardin.] 0161
21.If it rained here on this date 500 years ago and there’s no way to know whether it rained here on this date 500 years ago, then there are objective truths that we cannot know.
If it didn’t rain here on this date 500 years ago and there’s no way to know whether it rained here on this date 500 years ago, then there are objective truths that we cannot know.
There’s no way to know whether it rained here on this date 500 years ago.
∴ There are objective truths that we cannot know. [R, K, O]
22.If you know that you don’t exist, then you don’t exist.
If you know that you don’t exist, then you know some things.
If you know some things, then you exist.
∴ You exist. [K, E, S]
23.We have an idea of a perfect being.
If we have an idea of a perfect being, then this idea is either from the world or from a perfect being.
If this idea is from a perfect being, then there is a God.
∴ There is a God. [Use I, W, P, and G; from René Descartes, except for which dropped premise?]
24.The distance from A to B can be divided into an infinity of spatial points.
One can cross only one spatial point at a time.
If one can cross only one spatial point at a time, then one can’t cross an infinity of spatial points in a finite time.
If the distance from A to B can be divided into an infinity of spatial points and one can’t cross an infinity of spatial points in a finite time, then one can’t move from A to B in a finite time.
If motion is real, then one can move from A to B in a finite time.
∴ Motion isn’t real. [Use D, O, C, M, and R; from the ancient Greek Zeno of Elea, who denied the reality of motion.]
25.If the square root of 2 equals some fraction of positive whole numbers, then (we stipulate) the square root of 2 equals x/y and x/y is simplified as far as it can be.
If the square root of 2 equals x/y, then 2 = x2/y2.
If 2 = x2/y2, then 2y2 = x2.
If 2y2 = x2, then x is even.
If x is even and 2y2 = x2, then y is even.
If x is even and y is even, then x/y isn’t simplified as far as it can be.
∴ The square root of 2 doesn’t equal some fraction of positive whole numbers. [F, E, S, T, T′, X, Y]
7.3 Harder proofs
Our present proof strategy has four steps: START, S&I, RAA, and REFUTE. Some arguments require a further multiple-assumption ASSUME step. Here’s an example: 0162
· If the butler was at the party, then he fixed the drinks and poisoned the deceased.
· If the butler wasn’t at the party, then the detective would have seen him leave the mansion and would have reported this.
· The detective didn’t report this.
· ∴ The butler poisoned the deceased.
· (A ⊃ (F • P))
· (∼A ⊃ (S • R))
· ∼R
· ∴ P
START by assuming “∼P”:
· 1 (A ⊃ (F • P))
· 2 (∼A ⊃ (S • R))
· 3 ∼R
· [ ∴ P
· 4 asm: ∼P
Then we’re stuck. We can’t apply the S- or I-rules or RAA; and we don’t have enough simple wffs for a refutation. What can we do? On our newly expanded strategy, when we get stuck we’ll make another assumption. We pick a complex wff we haven’t used yet (1 or 2), pick left or right side, and assume it or its negation. Here we decide to assume the negation of the left side of line 1:
· 1 (A ⊃ (F • P))
· 2 (∼A ⊃ (S • R))
· 3 ∼R
· [ ∴ P
· 4 asm: ∼P
· 5 asm: ∼A {break 1}
We use S- and I-rules to derive further lines; but now we use two stars (one for each assumption). Lines 3 and 8 contradict:
· 1 (A ⊃ (F • P))
· 2 (∼A ⊃ (S • R))
· 3 ∼R
· [ ∴ P
· 4 asm: ∼P
· 5 asm: ∼A {break 1}
· ** 6 ∴ (S • R) {from 2 and 5}
· 7 ∴ S {from 6}
· 8 ∴ R {from 6}
Since we have a contradiction, we (1) block off the lines from the last assumption on down (this tells us not to use these lines, here 5 to 8, as we derive further lines and look for a contradiction), (2) derive the opposite of this last assumption, and (3) erase star strings with more stars than the number of remaining assumptions: 0163
Then we use S- and I-rules to derive further lines, and thus we get a second contradiction (lines 4 and 12):
Finally, we apply RAA again, this time on our original assumption:
To prove the argument valid, we need to get a contradiction for each assumption. We’ve accomplished this, and our proof is done. 0164
The most difficult part of multiple-assumption proofs is knowing when to make another assumption and what to assume.
(1) Make another assumption when you’re stuck. You may get that deep sense of confusion in your gut. More technically, being stuck means that you can’t apply S- or I-rules further – and yet you can’t prove the argument VALID (since you have no contradiction) or INVALID (since you don’t have enough simple wffs for a refutation). Don’t make additional assumptions too soon; it’s too soon if you can still apply S- or I-rules or RAA. Always use S- and I-rules and RAA to their limit before resorting to further assumptions.
(2) When you’re stuck, make an assumption that breaks a complex wff. Look for a complex wff that isn’t starred, blocked off, or broken (a wff is broken if we already have one side or its negation but not what we need to conclude anything new). This wff will have a NOT-BOTH, OR, or IF-THEN form:
∼(A • B)
(A ∨ B)
(A ⊃ B)
Assume either side or its negation. Here we could use any of these:
asm: A
asm: ∼A
asm: B
asm: ∼B
While any of the four works, our proof will go differently depending on which we use. Suppose we want to break “(A ⊃ B)”; compare what happens if we assume “A” or assume “∼A”:
(immediate gratification)
· (A ⊃ B)
· asm: A
· ∴ B
(delayed gratification)
· (A ⊃ B)
· asm: ∼A
· …
In the first case, we assume “A” and use an I-rule on “(A ⊃ B)” to get “B.” In the second case, we assume “∼A” and get nothing; but we may be able to use an I-rule on “(A ⊃ B)” later, after the “∼A” assumption dies (if it does) and we derive “A.” Delayed gratification tends to produce shorter proofs; it saves an average of one line, with the gain coming on invalid arguments. So sometimes a proof is simpler if you assume one thing rather than another.
Do the same with longer wffs. To break “((A • B) ⊃ (C • D)),” make any of these four assumptions:
asm: (A • B)
asm: ∼(A • B)
asm: (C • D)
asm: ∼(C • D)
Assume one side or its negation. Never assume the denial of a whole line.
Never make an assumption to break a wff that’s already broken. A wff is broken if we already have one side or its negation but not what we need to conclude anything new. So a “(A ⊃ B)” line, for example, is broken if we already have a not-blocked-off line with “∼A” or with “B.” In such a case, it won’t help us to make an assumption to break “(A ⊃ B).” 0165
After making our second assumption, we star the same things as before, but now we use more stars:
Use one star for each live assumption.
Star any wff simplified using an S-rule.
** (A • B)
––––––––
∴ A
∴ B
Star the longer wff used in an I-rule inference.
** (A ⊃ B)
A
––––––––
∴ B
A live assumption is one that isn’t blocked off. So if we have two live assumptions, then we use two stars. And if we have three live assumptions, then we use three stars. As before, starred lines are redundant; when doing a proof, focus on complex wffs that aren’t starred or blocked off and what can be derived from them. Multiple stars mean “You can ignore this line for now, but you may have to use it later.”
When we have multiple live assumptions and find a contradiction:
· block off the lines from the last live assumption on down (these lines are no longer to be used in the proof – since they depend on an assumption that we’ve concluded to be false);
· derive the opposite of this last assumption; and
· erase star strings with more stars than the number of remaining live assumptions (since the blocked-off lines that make these starred lines redundant are no longer available).
Note the part about erasing star strings with more stars than the number of remaining live assumptions. So if our second assumption dies, leaving us with just one live assumption, then we erase double-stars (“**”).
When our last live assumption leads to a contradiction, we’ve proved the argument to be valid. Valid arguments seldom require more than two assumptions. But if we get stuck again after making a second assumption, then we’ll need to make a third assumption.
Our final proof strategy can prove or refute any propositional argument (as we’ll show in §15.4):
1. START: Block off the conclusion and add “asm:” followed by the conclusion’s simpler contradictory.
2. S&I: Go through the complex wffs that aren’t starred or blocked off and use these to derive new wffs using S- and I-rules. Star (with one star for each live assumption) any wff you simplify using an S-rule, or the longer wff used in an I-rule inference. If you get a contradiction, then go to RAA (step 3). If you can’t derive anything further but there is a complex wff that isn’t starred or blocked off or broken, then go to ASSUME (step 4). If you can’t derive anything further and every complex wff is starred or blocked off or broken, then go to REFUTE (step 5).0166
3. RAA: Apply the RAA rule. If all assumptions are now blocked off, you’ve proved the argument valid. Otherwise, erase star strings having more stars than the number of live assumptions and return to step 2.
4. ASSUME: Pick a complex wff that isn’t starred or blocked off or broken. This wff will have one of these forms: “∼(A • B),” “(A ∨ B),” or “(A ⊃ B).” Assume one side or its negation and return to step 2.
5. REFUTE: Construct a refutation box containing any simple wffs (letters or their negation) that aren’t blocked off. In the original argument, mark each letter “1” or “0” or “?” depending on whether the box has the letter or its negation or neither. These truth conditions should make the premises all true and conclusion false – thus showing the argument to be invalid.
Let’s do another valid one (we’ll do invalid later). Here, after deriving a few lines, we get stuck and can’t go further. So we need to make another assumption. We could assume the left or right sides (or their denials) of lines 1, 2, or 4.
· 1 (A ⊃ (B • C))
· 2 (B ⊃ (A • C))
· [ ∴ ((A ∨ B) ⊃ C)
· * 3 asm: ∼((A ∨ B) ⊃ C)
· 4 ∴ (A ∨ B) {from 3}
· 5 ∴ ∼C {from 3}
We decide to assume the left side of line 1. Then we derive further lines to get a contradiction (5 and 9). We add double stars, since we have two live assumptions.
· ** 1 (A ⊃ (B • C))
· 2 (B ⊃ (A • C))
· [∴ ((A ∨ B) ⊃ C)
· * 3 asm: ∼((A ∨ B) ⊃ C)
· 4 ∴ (A ∨ B) {from 3}
· 5 ∴ ∼C {from 3}
· 6 asm: A {break 1}
· ** 7 ∴ (B • C) {from 1 and 6}
· 8 ∴ B {from 7}
· 9 ∴ C {from 7}
We then block off from assumption 6 down, conclude its opposite in line 10, and (since we now have only one live assumption) erase double stars. As we continue the proof, we ignore blocked-off lines (the original conclusion and 6 to 9).0167
We then derive further lines and get our second contradiction (lines 10 and 13). We apply RAA again, this time on our original assumption.
Since every assumption has led to a contradiction, our proof is done.
7.3a Exercise: LogiCola GHV
Prove each of these arguments to be valid (all are valid). 0168
· (B ∨ A)
· (B ⊃ A)
· ∴ ∼(A ⊃ ∼A)
· 1. (A ⊃ B)
(A ∨ (A • C))
∴ (A • B)
· 2. (((A • B) ⊃ C) ⊃ (D ⊃ E))
D
∴ (C ⊃ E)
· 3. (B ⊃ A)
∼(A • C)
(B ∨ C)
∴ (A ≡ B)
· 4. (A ∨ (D • E))
(A ⊃ (B • C))
∴ (D ∨ C)
· 5. ((A ⊃ B) ⊃ C)
(C ⊃ (D • E))
∴ (B ⊃ D)
· 6. (∼(A ∨ B) ⊃ (C ⊃ D))
(∼A • ∼D)
∴ (∼B ⊃ ∼C)
· 7. (∼A ≡ B)
∴ ∼(A ≡ B)
· 8. (A ⊃ (B • ∼C))
C
((D • ∼E) ∨ A)
∴ D
7.3b Exercise: LogiCola GHV
First appraise intuitively. Then translate into logic (using the letters given) and prove to be valid (all are valid).
1. Either the butler fixed the drink and poisoned the deceased, or the butler added poison later and poisoned the deceased.
If the butler poisoned the deceased, then the butler is guilty.
∴ The butler poisoned the deceased and is guilty. [Use F, P, A, and G.]
2. If I’m coming down with a cold and I exercise, then I’ll get worse and feel awful. If I don’t exercise, then I’ll suffer exercise deprivation and I’ll feel awful.
∴ If I’m coming down with a cold, then I’ll feel awful. [Use C, E, W, A, and D. This one is easier if you break premise 1 (not premise 2) to make your assumption.]
3. You’ll get an A if and only if you either get a hundred on the final exam or else bribe the teacher.
You won’t get a hundred on the final exam.
∴ You’ll get an A if and only if you bribe the teacher. [Use A, H, and B.] 0169
4. If President Nixon knew about the massive Watergate cover-up, then he lied to the American people on national television and he should resign.
If President Nixon didn’t know about the massive Watergate cover-up, then he was incompetently ignorant and he should resign.
∴ Nixon should resign. [K, L, R, I]
5. If you don’t compromise your principles, then you won’t get campaign money.
If you won’t get campaign money, then you won’t be elected.
If you compromise your principles, then you’ll appeal to more voters.
If you appeal to more voters, then you’ll be elected.
∴ You’ll be elected if and only if you compromise your principles. [C, M, E, A]
6. Moral judgments express either truth claims or feelings.
If moral judgments express truth claims, then “ought” expresses either a concept from sense experience or an objective concept that isn’t from sense experience.
“Ought” doesn’t express a concept from sense experience.
“Ought” doesn’t express an objective concept that isn’t from sense experience.
∴ Moral judgments express feelings and not truth claims. [T, F, S, O]
7. If Michigan either won or tied, then Michigan is going to the Rose Bowl and Gensler is happy.
∴ If Gensler isn’t happy, then Michigan didn’t tie. [W, T, R, H]
8. There are moral obligations.
If there are moral obligations and moral obligations are explainable, then either there’s an explanation besides God’s existence or else God’s existence would explain moral obligations.
God’s existence wouldn’t explain moral obligation.
∴ Either moral obligations aren’t explainable, or else there’s an explanation besides God’s existence. [M, E, B, G]
9. If determinism is true and Dr Freudlov correctly predicts (using deterministic laws) what I’ll do, then if she tells me her prediction I’ll do something else.
If Dr Freudlov tells me her prediction and yet I’ll do something else, then Dr Freudlov doesn’t correctly predict (using deterministic laws) what I’ll do.
∴ If determinism is true, then Dr Freudlov doesn’t correctly predict (using deterministic laws) what I’ll do or else she won’t tell me her prediction. [D, P, T, E]
10.If you make this demand on your son [that he leave Suzy or else not have his graduate schooling financed] and he leaves Suzy, then he’ll regret being forced to leave her and he’ll always resent you.
If you make this demand on your son and he doesn’t leave Suzy, then he’ll regret not going to graduate school and he’ll always resent you.
∴ If you make this demand on your son, then he’ll always resent you. [Use D, L, F, A, and G; this one is difficult.] 0170
7.4 Harder refutations
With multiple-assumption invalid arguments, we keep making assumptions until we get our refutation. Here’s an example:
· If the butler was at the party, he fixed the drinks and poisoned the deceased.
· If the butler wasn’t at the party, he was at a neighbor’s house.
· ∴ The butler poisoned the deceased.
· 1 (A0 ⊃ (F? • P0)) = 1 Invalid
· ** 2 (∼A0 ⊃ N1) = 1
· [ ∴ P0 = 0
· 3 asm: ∼P
· 4 asm: ∼A {break 1}
· 5 ∴ N {from 2 and 4}
N, ∼A, ∼P
We derive all we can and make additional assumptions when needed. We reach a refutation in which the butler was at a neighbor’s house, wasn’t at the party, and didn’t poison the deceased. This makes the premises all true and conclusion false.
Follow the five-step proof strategy of the previous section until you get a proof or a refutation. If every assumption leads to a contradiction, then you get a proof. But when do you know that the argument is invalid? When do you stop making further assumptions and instead construct a refutation box? Stop and refute when you can’t derive anything further (using S- or I-rules or RAA) and every complex wff is starred or blocked off or broken. (A complex wff is “broken” if we have one side or its negation but not what we need to conclude anything new.)
This invalid argument requires three assumptions:
· 1 (A0 ⊃ B?) = 1 Invalid
· 2 (C0 ⊃ D?) = 1
· 3 (F0 ⊃ (C0 • D?)) = 1
· [ ∴ (E1 ⊃ C0) = 0
· * 4 asm: ∼(E ⊃ C)
· 5 ∴ E {from 4}
· 6 ∴ ∼C {from 4}
· 7 asm: ∼A {break 1}
· 8 asm: ∼F {break 3}
E, ∼A, ∼C, ∼F
Here we can derive nothing further and all complex wffs are either starred (line 4), blocked off (original conclusion), or broken (lines 1–3). Our refutation, even without values for “B” or “D,” makes the premises all true and conclusion false.
Our proof strategy, if applied correctly, will always give a proof or refutation. How these go may depend on which lines we do first and what we decide to assume; proofs and refutations may differ but still be correct. 0171
7.4a Exercise: LogiCola GHI
Prove each of these arguments to be invalid (all are invalid).
· (A ∨ ∼(B ⊃ C))
· (D ⊃ (A ⊃ B))
· ∴(C ⊃ ∼(D ∨ A))
· 1 (A1 ∨ ∼ (B? ⊃ C1)) = 1 Invalid
· 2 (D0 ⊃ (A1 ⊃ B?)) = 1
· [ ∴ (C1 ⊃ ∼ (D0 ∨ A1)) = 0
· * 3 asm: ∼ (C ⊃ ∼ (D ∨ A))
· 4 ∴ C {from 3}
· 5 ∴ (D ∨ A) {from 3}
· 6 asm: A {break 1}
· 7 asm: ∼ D {break 2}
A, C, ∼D
1. ∼(A • B)
∴ (∼A • ∼B)
2. (A ⊃ ∼B)
∴ ∼(A ⊃ B)
3. (A ⊃ B)
(C ⊃ (∼D • E))
∴ (D ∨ F)
4. ∼(A • B)
∴ ∼(A ≡ B)
5. (A ⊃ (B • C))
((D ⊃ E) ⊃ A)
∴ (E ∨ C)
6. (∼A ∨ ∼B)
∴ ∼(A ∨ B)
7. ((A • B) ⊃ ∼(C • D))
C
(E ⊃ B)
∴ ∼E
8. (A ⊃ (B ⊃ C))
(B ∨ ∼(C ⊃ D))
∴ (D ⊃ ∼(A ∨ B))
7.4b Exercise: LogiCola G (HC & MC)
First appraise intuitively. Then translate into logic (using the letters given) and say whether valid (and give a proof) or invalid (and give a refutation).
1. If the maid prepared the drink, then the butler didn’t prepare it.
The maid didn’t prepare the drink.
If the butler prepared the drink, then he poisoned the drink and is guilty.
∴ The butler is guilty. [Use M, B, P, and G.]
2. If you tell your teacher that you like logic, then your teacher will think that you’re insincere and you’ll be in trouble.
If you don’t tell your teacher that you like logic, then your teacher will think that you dislike logic and you’ll be in trouble.
∴ You’ll be in trouble. [Use L, I, T, and D.]
3. If we don’t get reinforcements, then the enemy will overwhelm us and we won’t survive.
∴ If we do get reinforcements, then we’ll conquer the enemy and we’ll survive. [Use R, O, S, and C.] 0172
4. If Socrates didn’t approve of the laws of Athens, then he would have left Athens or would have tried to change the laws.
If Socrates didn’t leave Athens and didn’t try to change the laws, then he agreed to obey the laws.
Socrates didn’t leave Athens.
∴ If Socrates didn’t try to change the laws, then he approved of the laws and agreed to obey them. [Use A, L, C, and O; from Plato’s Crito, which argued that Socrates shouldn’t disobey the law by escaping from jail.]
5. If I hike the Appalachian Trail and go during late spring, then I’ll get maximum daylight and maximum mosquitoes.
If I get maximum mosquitoes, then I won’t be comfortable.
If I go right after school, then I’ll go during late spring.
∴ If I hike the Appalachian Trail and don’t go right after school, then I’ll be comfortable. [A, L, D, M, C, S]
6. [Logical positivism says “Every genuine truth claim is either experimentally testable or true by definition.” This view, while once popular, is self-refuting and hence not very popular today.]
If LP (logical positivism) is true and is a genuine truth claim, then it’s either experimentally testable or true by definition.
LP isn’t experimentally testable.
LP isn’t true by definition.
If LP isn’t a genuine truth claim, then it’s not true.
∴ LP isn’t true. [T, G, E, D]
7. If you give a test, then students either do well or do poorly.
If students do well, then you think you made the test too easy and you’re frustrated.
If students do poorly, then you think they didn’t learn any logic and you’re frustrated.
∴ If you give a test, then you’re frustrated. [Use T, W, P, E, F, and L; from a class who tried to talk me out of giving a test.]
8. If the world contains moral goodness, then the world contains free creatures and the free creatures sometimes do wrong.
If the free creatures sometimes do wrong, then the world is imperfect and the creator is imperfect.
∴ If the world doesn’t contain moral goodness, then the creator is imperfect. [M, F, S, W, C]
9. We’ll find your action’s cause, if and only if your action has a cause and we look hard enough.
If all events have causes, then your action has a cause.
All events have causes.
∴ We’ll find your action’s cause, if and only if we look hard enough. [F, H, L, A]0173
10.Herman sees that the piece of chalk is white.
The piece of chalk is the smallest thing on the desk.
Herman doesn’t see that the smallest thing on the desk is white. (He can’t see the whole desk and so can’t tell that the piece of chalk is the smallest thing on it.)
If Herman sees a material thing, then if he sees that the piece of chalk is white and the piece of chalk is the smallest thing on the desk, then he sees that the smallest thing on the desk is white.
If Herman doesn’t see a material thing, then he sees a sense datum.
∴ Herman doesn’t see a material thing, but he does see a sense datum. [Use H, P, H′, M, and S. This argument attacks direct realism: that we directly perceive material objects and not just sensations.]
11.If the final capacitor in the transmitter is arcing, then the SWR (standing wave ratio) is too high and the efficiency is lowered.
If you hear a cracking sound, then the final capacitor in the transmitter is arcing.
∴ If you don’t hear a cracking sound, then the SWR isn’t too high. [A, H, L, C]
12.If we can know that God exists, then we can know God by experience or we can know God by logical inference from experience.
If we can’t know God empirically, then we can’t know God by experience and we can’t know God by logical inference from experience.
If we can know God empirically, then “God exists” is a scientific hypothesis and is empirically falsifiable.
“God exists” isn’t empirically falsifiable.
∴ We can’t know that God exists. [K, E, L, M, S, F]
13.If I perceive, then my perception is either delusive or veridical.
If my perception is delusive, then I don’t directly perceive a material object.
If my perception is veridical and I directly perceive a material object, then my experience in veridical perception would always differ qualitatively from my experience in delusive perception.
My experience in veridical perception doesn’t always differ qualitatively from my experience in delusive perception.
If I perceive and I don’t directly perceive a material object, then I directly perceive a sensation.
∴ If I perceive, then I directly perceive a sensation and I don’t directly perceive a material object. [Use P, D, V, M, Q, and S. This argument from illusion attacks direct realism: that we directly perceive material objects and not just sensations.]
14.If you’re romantic and you’re Italian, then Juliet will fall in love with you and will want to marry you.
If you’re Italian, then you’re romantic.
∴ If you’re Italian, then Juliet will want to marry you. [R, I, F, M]
15.If emotions can rest on factual errors and factual errors can be criticized, then we can criticize emotions.
If we can criticize emotions and moral judgments are based on emotions, then beliefs about morality can be criticized and morality isn’t entirely non-rational.
∴ If morality is entirely non-rational, then emotions can’t rest on factual errors. [E, F, W, M, B, N] 0174
7.5 Copi proofs
There are many proof methods for propositional logic. Copi proofs are based on an early and still popular method. Copi proofs use a somewhat standard set of inference and replacement rules.1 These eight inference rules, like our S- and I-rules, let us infer whole lines from previous whole lines (here each capital letter may be uniformly replaced by any wff):
1 This proof method goes back to Irving Copi’s Introduction to Logic (New York: Macmillan, 1953) and has appeared with variations in many books. Copi’s original list (his p. 259) also had Destructive Dilemma (“((P ⊃ Q) • (R ⊃ S)), (∼Q ∨ ∼S) ∴ (∼P ∨ ∼R)”) but omitted Repetition’s second part. Absorption (“(P ⊃ Q) ∴ (P ⊃ (P • Q))” and “(P ⊃ (P • Q)) ∴ (P ⊃ Q)”) was sometimes added later. I simplified some names and gave each rule a two-letter abbreviation.
AD Addition
P
––––––
(P ∨ Q)
CJ Conjunction
P Q
–––––––
(P • Q)
DI Dilemma
((P ⊃ Q) • (R ⊃ S)) (P ∨ R)
–––––––––––––––
(Q ∨ S)
DS Disjunctive Syllogism
(P ∨ Q) ∼P
–––––––
Q
HS Hypothetical Syllogism
(P ⊃ Q) (Q ⊃ R)
–––––––
(P ⊃ R)
MP Modus Ponens
(P ⊃ Q) P
–––––––
Q
MT Modus Tollens
(P ⊃ Q) ∼Q
–––––––
∼P
SP Simplification
(P • Q)
–––––––
P
To explore how these work, we’ll compare them to our S- and I-rules.
Our first three S-rules are AND, NOR, and NIF:
|
AND |
NOR |
NIF |
|
(P • Q) –––––– P, Q |
~(P ∨ Q) –––––– ~P, ~Q |
~(P ⊃ Q) –––––– P, ~Q |
Copi can derive the AND rule. We can get the left side of “(P • Q)” by using SP (Simplification) directly on line 1, as indicated by the “{SP 1}” justification:
1. (P • Q)
2. P {SP 1}
To get the right side, we first switch sides to get “(Q • P),” using replacement rule CM (Commutation), which we’ll present later. Then we use SP:
1. (P • Q)
2. (Q • P) {CM 1)
3. Q (SP 2} 0175
Copi can also derive the NOR conclusions. To get the left side, we first apply the DM (De Morgan) replacement rule, which we’ll present later, to go from “∼(P ∨ Q)” to “(∼P • ∼Q).” Then we use SP to derive “∼P”:
1. ∼(P ∨ Q)
2. (∼P • ∼Q) {DM 1}
3. ∼P {SP 2}
To get the right side, we use similar reasoning, but we have to again switch sides using CM:
1. ∼(P ∨ Q)
2. (∼P • ∼Q) {DM 1}
3. (∼Q • ∼P) {CM 2}
4. ∼Q {SP 3}
Deriving NIF is more involved. To get the left side, we first reshape the IF-THEN into an OR (using Implication replacement rule IM) and then an AND (using De Morgan replacement rule DM). We apply SP to get “∼∼P,” and then the double-negation replacement rule DN to get “P”:
1. ∼(P ⊃ Q)
2. ∼(∼P ∨ Q) {IM 1}
3. (∼∼P • ∼Q) {DM 2}
4. ∼∼P {SP 3}
5. P {DN 4}
Getting the right side is similar, but we switch sides using CM before using SP to get “∼Q”:
1. ∼(P ⊃ Q)
2. ∼(∼P ∨ Q) {IM 1}
3. (∼∼P • ∼Q) {DM 2}
4. (∼Q • ∼∼P) {CM 3}
5. ∼Q {SP 4}
These examples show how difficult the Copi method can be to use. But that’s the challenge – it makes us think hard about how to derive a conclusion, and maybe think out various possible approaches first; some teachers like Copi proofs for exactly this reason.
Replacement rules are important in Copi proofs. These ten replacement rules let you switch one occurrence of identical formulas anywhere in a wff:
AS Association
(P ∨ (Q ∨ R)) = ((P ∨ Q) ∨ R)
(P • (Q • R )) = ((P • Q) • R)
CM Commutation
(P ∨ Q) = (Q ∨ P)
(P • Q) = (Q • P)
DB Distribution
(P • (Q ∨ R)) = ((P • Q) ∨ (P • R))
(P ∨ (Q • R)) = ((P ∨ Q) • (P ∨ R))
DM De Morgan
∼ (P • Q) = (∼P ∨ ∼Q))
∼(P ∨ Q) = (∼P • ∼Q)
DN Double Negation
P = ∼∼P
EQ Equivalence
(P ≡ Q) = ((P ⊃ Q) • (Q ⊃ P))
(P ≡ Q) = ((P • Q) ∨ (∼P • ∼Q))
EX Exportation
((P • Q) ⊃ R) = (P ⊃ (Q ⊃ R))
IM Implication
(P ⊃ Q) = ( ∼P ∨ Q)
0176
RP Repetition
P = (P ∨ P)
P = (P • P)
TR Transposition
(P ⊃ Q) = (∼Q ⊃ ∼P)
These reshape formulas to fit the inference rules.
Let’s consider how the Copi method can mirror our I-rules:
OR conclusions are easy to derive using DS (Disjunctive Syllogism); for the second version, we also switch sides using CM:
1. (P ∨ Q)
2. ∼P
3. Q {DS 1+2}
1. (P ∨ Q)
2. ∼Q
3. (Q ∨ P) {CM 1}
4. P {DS 2+3}
NOT-BOTH uses DS with DM (De Morgan):
1. ∼(P • Q)
2. P
3. (∼P ∨ ∼Q) {DM 1}
4. ∼∼P {DN 2}
5. ∼Q {DS 3+4}
1. ∼(P • Q)
2. Q
3. (∼P ∨ ∼Q) {DM 1}
4. (∼Q ∨ ∼P) {CM 3}
5. ∼∼Q {DN 2}
6. P {DS 4+5}
Since Copi rules take “not” very strictly, we can’t on the left go directly from “P” and “(∼P ∨ ∼Q)” to get “∼Q”; instead, we have to double negate “P” to get “∼∼P,” which is like the first part but starts with an additional squiggle. MP (Modus Ponens) and MT (Modus Tollens) parallel our IF-THEN forms.
Here’s a Copi proof for the butler example in §7.1:
Conclusion: B
1. T
2. (T ⊃ (B ∨ M))
3. (M ⊃ H)
4. ∼H
5. (B ∨ M) {MP 1+2}
6. ∼M {MT 3+4}
7. (M ∨ B) {CM 5}
8. B {DS 6+7} 0177
If we try to prove the invalid butler example in §7.2, we won’t derive the conclusion; but this may just be due to our lack of ingenuity. The Copi method won’t show invalid arguments to be invalid; so it’s normally used only on arguments already known to be valid, which limits the method’s usefulness.
Conclusion: B
· T
· (T ⊃ (B ∨ M))
· (M ⊃ H)
· (B ∨ M) {MP 1+2}
· ? ? ?
Here’s a Copi proof for the multiple-assumption butler example in §7.3:
Conclusion: P
1. (A ⊃ (F • P))
2. (∼A ⊃ (S • R))
3. ∼R
4. (∼R ∨ ∼S) {AD 3}
5. ∼(R • S) {DM 4}
6. ∼(S • R) {CM 5}
7. ∼∼A {MT 2+6}
8. A {DN 7}
9. (F • P) {MP 1+8}
10.(P • F) {CM 9}
11.P {SP 10}
Lines 4 to 6 use a common strategy: think of what wff we need – here we need “∼(S • R)” to use with line 2 and MT – and how to get it from what we have – here “∼R” can provide “(∼R ∨ ∼S),” which we reshape into “∼(S • R).”
So far, we’ve used Copi direct proofs, where the conclusion is derived from the premises without making any assumptions. Copi also provides for conditional proofs and indirect proofs [reductio ad absurdum] (using CP and RA):
CP Conditional Proof
If you assume P and later derive Q, then you can star all the lines from P to Q [showing that you aren’t to use them to derive further steps] and then derive (P ⊃ Q).
RA Reductio ad Absurdum
If you assume P and later derive (Q • ∼Q), then you can star all the lines from P to (Q • ∼Q) [showing that you aren’t to use them to derive further steps] and then derive ∼P.
The proof isn’t done until all assumptions are starred. Here are examples (add “*” when applying RA or CP; ignore starred lines in deriving further steps):
Conclusion: ((A • B) ⊃ A)
1. (A • B) {Assume} *
2. A {SP 1} *
3. ((A • B) ⊃ A) {CP 1+2}
Conclusion: (A ∨ ∼A)
1. ∼(A ∨ ∼A) {Assume} *
2. (∼A • ∼∼A) {DM 1} *
3. ∼∼(A ∨ ∼A) {RA 1+2}
4. (A ∨ ∼A) {DN 3} 0178
CP and RA are useful in proving logical truths from zero premises. CP is convenient for proving conditional conclusions. And if you’re really confused on how to do a problem, I suggest that you start by assuming the conclusion’s opposite; try to derive a contradiction and then apply RA.
Comparing our method to Copi’s, all the same arguments are provable. Our method is easier to learn (with a smaller and more systematic set of rules), easier to use (with a proof procedure that doesn’t require guesswork or intuition), and more powerful (since it can refute invalid arguments). But you might want to learn the Copi method too; Copi proofs are good mental exercise and can be fun (especially on LogiCola) – and Copi rules are sometimes assumed in philosophical discussions.
On LogiCola, you do Copi proofs by picking “Copi Proofs” and the level of difficulty (Easier / Harder / Mixed); you get the same randomly generated problems (but only valid ones) as with our usual proofs. You repeatedly type the next wff, click (or touch) the inference rule, and then click (or touch) the previous wffs from which your step follows. There are no arrows to get the next line or finish the problem; but you can quit the problem (which costs you points) or paste your own problems (or ones from your teacher). You can also copy previous lines or the conclusion into the answer space, so you can modify them to give your next line. While Copi proofs are difficult, you’ll soon get the hang of it.
7.5a and 7.5b Exercise: LogiCola GEO
Do Copi proofs for problems in §§7.1a and 7.1b (all are valid). These are easier problems.
7.5c and 7.5d Exercise: LogiCola GHO and GMO
Do Copi proofs for problems in §§7.3a and 7.3b (all are valid). These are harder problems.
7.6 Truth trees
Also common are truth trees, which break formulas into the cases that make them true. Here’s a truth tree for “(A • ∼B), (B ∨ C) ∴ C” – which comes out as valid, because every branch closes:
0179 First write the premises and the conclusion’s contradictory. Then break the complex formulas into the cases that make them true, to see if there’s some way to get premises all true and conclusion false. Simplify “(A • ∼B)” into “A” and “∼B” and then star it (as broken). Branch “(B ∨ C)” into the two cases that make it true and then star it (as broken); so one branch has “B” and another has “C.” Both branches are self-contradictory, since the first has “B” on the branch and “∼B” on the trunk – and the second has “C” on the branch and “∼C” on the trunk; close both branches by adding “*” to the bottom. The argument is VALID, since having premises all true and conclusion false is impossible.
Truth trees use simplifying and branching rules (both apply only to whole lines). These simplifying rules simplify a formula into smaller parts:
When you use these, put a star after the original formula to show that it’s broken (this means that its truth is assured by the truth of some smaller parts below). Use simplifying rules before branching rules (this is more efficient).
These branching rules branch a formula into the two sub-cases that would make it true (so “∼(P • Q)” is true just if “∼P” is true or “∼Q” is true):
When you use these, put a star after the original formula to show that it’s broken; add the sub-branches to the bottom of any bottom branch further down that isn’t yet marked as closed (self-contradictory).
To test an argument, write the premises and the conclusion’s contradictory. Keep applying the simplifying and branching rules to complex unstarred formulas. Close a branch when it has contradictory formulas; a closed branch is a failed attempt to make the premises all true and conclusion false. If all branches close, then the argument is valid. If some branch doesn’t close and yet all its complex wffs are starred (broken), then there’s a possible way to get premises all true and conclusion false, and so the argument is invalid. 0180
To show further how this works, I’ll do truth trees for the butler examples in §§7.1–7.4. The argument in §7.1 comes out valid, since every branch closes:
Here we branch line 2 into “∼T” and “(B ∨ M),” branch the latter into “B” and “M,” and branch “(M ⊃ H)” into “∼M” and “H”; every branch closes, and so it’s valid. The butler argument in §7.2 is invalid, since without “∼H” the branch ending in “H” doesn’t close (the refutation has: T, M, H, and ∼B):
The butler argument in §7.3 comes out valid, since every branch closes:
0181 The butler argument in §7.4 comes out invalid, since the branch ending in “N” doesn’t close (the refutation has: N, ∼A, and ∼P):
As compared with Copi proofs, truth trees are easier to do (since they use an easily learned strategy) and can test for validity or invalidity. But truth trees don’t mirror ordinary reasoning as well; they give a mechanical way to test validity instead of a way to help develop reasoning skills. Our proof method tries to combine the strengths of both methods. Like truth trees, our proofs have an easily learned strategy, keep breaking formulas into simpler parts, and can test for validity or invalidity. But like Copi proofs, our proofs give a linear derivation of formulas that somewhat mirrors ordinary reasoning. Our proofs use similar simplification rules as truth trees, but replace branching with inference rules and assumptions.
On LogiCola, you do truth trees by picking “Treez” and the level of difficulty (Easier / Harder / Mixed); you get the same randomly generated problems (valid and invalid) as with regular proofs. You do this exercise entirely by clicking or touching (no typing); follow the directions at the bottom. The “program closes branches” option automatically closes self-contradictory branches, while “you close branches” has you close these yourself (but without losing points for errors). The “automatic double-negation” option simplifies double negations automatically (so “∼(A ∨ ∼B)” simplifies into “∼A” and “B” – instead of “∼A” and “∼∼B”). You can click (or touch) arrows at the top to give you the next line or finish the problem (but without getting credit for the problem); and you can use these arrows to step through sample proofs before doing them on your own.
7.6a Exercise: LogiCola GEZ
Do truth trees for problems in §§7.1a, 7.1b, 7.2a, and 7.2b. These are easier problems.
7.6b Exercise: LogiCola GHZ and GMZ
Do truth trees for problems in §§7.3a, 7.3b, 7.4a, and 7.4b. These are harder problems.