PHIL12A
Section answers, 28 Feb 2011
Julian Jonker
1
How much do you know?
Give formal proofs for the following arguments.
1. (Ex 6.18)
1
A∨B
2
A ∨ ¬¬B
Proof:
1
A∨B
2
A
3
A ∨ ¬¬B
4
B
∨Elim: 2
5
¬B
6
⊥
⊥Intro: 4,5
7
¬¬B
¬Intro: 5-6
8
A ∨ ¬¬B
∨Intro: 7
9
A ∨ ¬¬B
∨Elim: 1, 2-3, 4-8
1
2. (Ex 6.25)
1
¬A ∧ ¬B
2
¬(A ∨ B)
Proof:
1
¬A ∧ ¬B
2
A∨B
3
A
4
¬A
∧Elim: 1
5
⊥
⊥Intro: 3,4
6
B
7
¬B
∧Elim: 1
8
⊥
⊥Intro: 6,7
9
10
⊥
∨Elim: 2, 3-5, 6-8
¬(A ∨ B)
2
¬Intro: 2, 3-9
3. (Ex 6.27)
1
(A ∧ B) ∨ (C ∧ D)
2
(B ∧ C) ∨ (D ∧ E)
3
C ∨ (A ∧ E)
Proof:
1
(A ∧ B) ∨ (C ∧ D)
2
(B ∧ C) ∨ (D ∧ E)
3
(A ∧ B)
4
A
∧Elim: 3
5
(B ∧ C)
6
C
∧Elim: 5
7
C ∨ (A ∧ E)
∨Intro: 6
8
D∧E
9
E
∧Elim: 8
10
A∧E
∧Intro: 4, 9
11
C ∨ (A ∧ E)
∨Intro: 10
12
C ∨ (A ∧ E)
13
C ∧D
14
C
∧Elim: 13
15
C ∨ (A ∧ E)
∨Intro: 14
16
∨Elim: 2, 5-7, 8-11
C ∨ (A ∧ E)
∨Elim: 1, 3-12, 13-15
3
4.
1
A ∨ (B ∧ (C ∨ D))
2
(A ∨ (B ∧ C)) ∨ (A ∨ (B ∧ D))
Proof:
1
A ∨ (B ∧ (C ∨ D))
2
A
3
A ∨ (B ∧ C)
∨Intro: 2
4
(A ∨ (B ∧ C)) ∨ (A ∨ (B ∧ D))
∨Intro: 3
5
B ∧ (C ∨ D)
6
C ∨D
7
C
8
B
∧Elim: 5
9
B∧C
∧Intro: 7, 8
10
A ∨ (B ∧ C)
∨Intro: 9
11
(A ∨ (B ∧ C)) ∨ (A ∨ (B ∧ D))
∨Intro: 10
12
D
13
B
∧Elim: 5
14
B∧D
∧Intro: 12, 13
15
A ∨ (B ∧ D)
∨Intro: 14
16
(A ∨ (B ∧ C)) ∨ (A ∨ (B ∧ D))
∨Intro: 15
17
18
∧Elim: 5
(A ∨ (B ∧ C)) ∨ (A ∨ (B ∧ D))
(A ∨ (B ∧ C)) ∨ (A ∨ (B ∧ D))
4
∨Elim: 6, 7-11, 12-16
∨Elim: 1, 2-4, 5-17
5.
1
A ∨ (B ∨ C)
2
¬(A ∨ B)
3
¬(¬C ∧ D)
Proof:
1
A ∨ (B ∨ C)
2
¬(A ∨ B)
3
¬C ∧ D
4
A
5
A∨B
∨Intro: 4
6
⊥
⊥Intro: 2, 5
7
B∨C
8
B
9
A∨B
∨Intro: 8
10
⊥
⊥Intro: 2, 9
11
C
12
¬C
∧Elim: 3
13
⊥
⊥Intro: 11, 12
⊥
14
15
16
∨Elim: 7, 8-10, 11-13
⊥
∨Elim: 1, 4-6, 7-14
¬(¬C ∧ D)
¬Intro: 3-15
5
2
Challenge question
The elimination and introduction rules for the Fitch system of formal proof all seem pretty logical, and can be
justified on the basis of the truth-functional descriptions of the connectives. There are two questions you might
ask about them, though: (a) if you prove something in Fitch, how are you guaranteed that the argument thus
proved is valid? (b) if you know an arbitrary argument is valid, can you find a proof for it in Fitch?
The textbook eventually covers these questions, which are questions about the soundness and completeness
of a proof system. (NB do not confuse soundness of a proof system with soundness of an argument!) But you
can already prove soundness for the rules we have. What you are proving, precisely, is the following:
Theorem 1. If P1 , ..., Pn are premises, and S is a sentence that can be formally derived from them, then S is a
tautological consequence of P1 , ..., Pn .
Here are guidelines for the proof:
• Think about what a formal proof is. It is a sequence of sentences, each sentence bearing a particular
relation to prior sentences in the sequence. S is just the final such step. So if we could prove that every
step in the proof is a tautological consequence of the premises, we would have proved the claim.
• It might seem harder to show that the above property belongs to every sentence in a proof. But not if we
use proof by contradiction. Suppose our proof system is not sound. Then there is some proof for which
the conclusion S is not a tautological consequence of premises P1 , ..., Pn . And this in turn means that
some step in the proof went wrong, that is, that there is a sentence in the proof that is not a tautological
consequence of the premises.
• Suppose there is such a sentence, call it Q. Q is derived from some prior sentences by one of our inference
rules. So now we have a proof by cases. In each case suppose one of the inference rules was responsible
for the bad step. If you can show, in each case, that this leads to contradiction, then you have proved
that we are led to contradiction by our assumption that there is a sentence Q that is not a tautological
consequence of P1 , ..., Pn .
Proof. Consider an arbitrary formal proof p that deduces conclusion S from premises P1 , ..., Pn , and is comprised of
a finite number of steps, each step being either an arbitrary hypothesis or a sentence that follows from prior steps in
the proof and the application of one of the inference rules Reit, ∧Intro, ∧Elim, ∨Intro, ∨Elim, ¬Intro, ¬Elim, ⊥Intro,
or ⊥Elim.
We distinguish between the assumptions that are in force at each step of the proof, and sentences that follow from
the assumptions in force at each step. At each step the assumptions in force include the premises P1 , ..., Pn , as well
any hypotheses that have been assumed for any subproof open at that step.
Assume, for the sake of proof by contradiction, that S is not a tautological consequence of P1 , ..., Pn . This means
that some step in the proof is not a tautological consequence of the assumptions in force at that step. (Otherwise truth
would be preserved at each step, and so the truth of the premises would ensure the truth of the conclusion.)
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Consider the first such invalid step, establishing sentence R. We can easily show, using proof by cases, that none of
our rules of inference could have resulted in this invalid step. I will just consider one case (but you should do the others
as an exercise): ∨Elim. Suppose that the invalid step results from the application of this rule to some prior sentence
Q1 ∨ Q2 ∨ ... ∨ Qj and j subproofs, each starting with a distinct disjunct of Q1 ∨ Q2 ∨ ... ∨ Qj and ending in R, and
that assumptions A1 , ..., Ai are in force at this step. Since the step we are currently considering is the first invalid step,
we know that in each subproof R is a tautological consequence of the relevant disjunct and the assumptions A1 , ..., Ai .
Now consider the truth table for assumptions A1 , ..., Ai , Q1 ∨ Q2 ∨ ... ∨ Qj , disjuncts Q1 , ..., Qj and R. Since,
by assumption, R is not a tautological consequence of Q1 ∨ Q2 ∨ ... ∨ Qj and A1 , ..., Ai , there is some row in which
A1 , ..., Ai and Q1 ∨ Q2 ∨ ... ∨ Qj are all true but R comes out false. Yet for Q1 ∨ Q2 ∨ ... ∨ Qj to be true, one of the
disjuncts – call it Qk – must be true. But this is a contradiction, since we know that R is a tautological consequence
of Qk , and so R must be true in this row of the truth table.
Therefore our assumption that S is not a tautological consequence of P1 , ..., Pn must be wrong.
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