Please complete all questions on the problem set.I will attach lecture slides for reference
12a_spring_2019_problem_set_7.pdf
12a_session_15_march_12_2019.pdf
12a_session_16_march_14_2019.pdf
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PHIL 12A – Spring 2019
Problem Set 7
65 points
Reductio Ad Absurdum
1. (10 points) Recall that π is an irrational number. Outline a natural deduction proof
that formalizes the reductio step in the following proof.
Theorem 1. There is some digit among 1, 2, . . . , 9 that occurs infinitely many times
in the decimal expansion of π.
Proof. Suppose it is not the case that there is some digit among 1, 2, . . . , 9 that occurs
infinitely many times in the decimal expansion of π. Hence each occurs only finitely
many times. Then since there are only finitely many digits among 1, 2, . . . , 9, and
each occurs only finitely many times in the decimal expansion of π, it follows that the
decimal expansion of π is a finite sequence of those digits (followed by only 0’s). This
contradicts the fact that π is irrational. Thus, we conclude that there is some digit
among 1, 2, . . . , 9 that occurs infinitely many times in the decimal expansion of π.
However, as of 2019, no one knows which digit(s) occur infinitely many times in the
decimal expansion of π.
2. (10 points) Give natural deduction proofs of the following formulas from the given
assumptions:
(a) p from assumption ¬(p → q);
(b) ¬¬p → p from no assumptions.
3. (5 points) Extra credit. Give a natural deduction proof of ¬p ↔ q from assumption
¬(p ↔ q).
4. (10 points) Prove that any application of EFQ can be replaced by applications of
Reiteration and RAA. Thus, EFQ becomes redundant when we add RAA.
Disjunction Introduction
5. (10 points) Give natural deduction proofs of the following formulas from the given
assumptions:
(a) ¬p ∧ ¬q from assumption ¬(p ∨ q);
(b) ¬p ∨ ¬q from assumption ¬(p ∧ q) (hint: use RAA).
1
Disjunction Elimination
6. (10 points) A prime number is a positive integer greater than 1 that is divisible without
remainder only by itself and 1. A set is infinite if and only if it is not finite.
Outline a natural deduction proof that formalizes the negation introduction step and
the proof by cases in the proof of Theorem 3 below.
Lemma 1. Let a, b, and c be integers. If a divides b without remainder and a divides
c without remain, then a divides (b − c) without remainder.
Theorem 2 (Fundamental Theorem of Arithmetic). Every integer greater than 1 is
a prime number or a product of prime numbers.1
Theorem 3 (Euclid’s Theorem). The set of prime numbers is infinite.
Proof. Suppose that the set of prime numbers is finite. List all the primes as p1 , . . . , pn
and let
p
= p1 × · · · × pn
q
= p + 1.
Either q is prime or it is not prime, so we can reason by cases.
Case 1: suppose q is prime. Then since q is greater than each of p1 , . . . , pn , we have a
contradiction with our assumption that p1 , . . . , pn are all the primes.
Case 2: suppose q is not prime. Then by Theorem 2, q is divisible without remainder
by some prime—hence by some pi , as p1 , . . . , pn are all the primes. Since pi divides q
without remainder and pi clearly divides p without remainder, it follows by Lemma 1
that pi divides q − p = 1 without remainder. But since pi is prime, it is greater than
1, so it does not divide 1 without remainder. This is a contradiction.
In either case, we obtained a contradiction, so the set of prime numbers is infinite.
7. (15 points) Give natural deduction proofs of the following formulas:
(a) p ∨ q from assumption q ∨ p;
(b) r from assumptions p ∨ q, p → r, and q → r;
(c) q from assumptions p ∨ q and ¬p.
1 Moreover,
its representation as a product of prime numbers is unique modulo the order of the factors.
2
Natural Deduction X:
Disjunction Introduction
PHIL 12A – Introduction to Logic
with Wes Holliday
University of California, Berkeley
Proving a Disjunction
One way to prove a disjunction j _ y is to prove j.
Another way to prove a disjunction j _ y is to prove y.
Proving a Disjunction
An integer n is even if n = 2k for some integer k.
An integer n is odd if n = 2k + 1 for some integer k.
Proving a Disjunction
An integer n is even if n = 2k for some integer k.
An integer n is odd if n = 2k + 1 for some integer k.
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proving a Disjunction
An integer n is even if n = 2k for some integer k.
An integer n is odd if n = 2k + 1 for some integer k.
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
Proving a Disjunction
An integer n is even if n = 2k for some integer k.
An integer n is odd if n = 2k + 1 for some integer k.
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
Proving a Disjunction
An integer n is even if n = 2k for some integer k.
An integer n is odd if n = 2k + 1 for some integer k.
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
For the inductive step, . . . to be continued . . .
Proving a Disjunction
An integer n is even if n = 2k for some integer k.
An integer n is odd if n = 2k + 1 for some integer k.
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
For the inductive step, . . . to be continued . . .
We can represent the logical
..
.
move in the base case as follows:
..
.
i
0 is even
i +1
0 is even _ 0 is odd
_I, i
Disjunction Introduction
_ Introduction states that if you have a correct partial proof in which j
appears or y appears, then you may correctly add j _ y on a new line in the
same column (of the same subproof):
..
.
..
.
..
.
..
.
i
..
.
j
..
.
i
..
.
y
..
.
j
j_y
j
j_y
_I, i
_I, i
Proving Disjunctions Nonconstructively
Proving a disjunct is not the only way to prove a disjunction.
Proving Disjunctions Nonconstructively
Proving a disjunct is not the only way to prove a disjunction.
Reductio Ad Absurdum allows one to prove a disjunction by assuming its
negation and then deriving a contradiction.
Proving Disjunctions Nonconstructively
Proving a disjunct is not the only way to prove a disjunction.
Reductio Ad Absurdum allows one to prove a disjunction by assuming its
negation and then deriving a contradiction.
In this way, one can prove disjunctions without have any idea which disjunct
holds—just as RAA lets one prove that there are irrational a, b such that ab is
rational without proving that any particular a, b work.
Ex.
1
n
p_ p
Ex.
1
n´1
n
(p _ p )
j^
p_ p
j
RAA, 1–n ´ 1
Ex.
1
(p _ p )
n´1
p^
n
p_ p
p
RAA, 1–n ´ 1
Ex.
1
(p _ p )
i
p
n´2
p
n´1
p^
n
p_ p
p
^I, i, n ´ 2
RAA, 1–n ´ 1
Ex.
1
(p _ p )
2
p
i ´1
j^
i
I, 2–i ´ 1
p
n´2
p
n´1
p^
n
j
p_ p
p
^I, i, n ´ 2
RAA, 1–n ´ 1
Ex.
1
(p _ p )
2
p
3
p_ p
i ´1
j^
i
j
I, 2–i ´ 1
p
n´2
p
n´1
p^
n
p_ p
_I, 2
p
^I, i, n ´ 2
RAA, 1–n ´ 1
Ex.
1
(p _ p )
2
p
3
p_ p
(p _ p )
4
(p _ p ) ^ (p _ p )
5
6
p
p
n´1
p^
p_ p
R, 1
^I, 3, 4
I, 2–5
n´2
n
_I, 2
p
^I, 6, n ´ 2
RAA, 1–n ´ 1
Ex.
1
(p _ p )
2
p
3
p_ p
(p _ p )
4
(p _ p ) ^ (p _ p )
5
6
p
R, 1
^I, 3, 4
I, 2–5
7
p
8
analogous to 3–5
9
p
10
p^
11
_I, 2
p_ p
I, 7–8
p
^I, 6, 9
RAA, 1–10
Disjunction Property
If we drop RAA from our proof system, then the resulting proof system has the
DISJUNCTION PROPERTY:
Disjunction Property
If we drop RAA from our proof system, then the resulting proof system has the
DISJUNCTION PROPERTY:
§
if there is a proof of j _ y (from no assumptions), then there is a proof of
j or there is a proof of y.
Disjunction Property
If we drop RAA from our proof system, then the resulting proof system has the
DISJUNCTION PROPERTY:
§
if there is a proof of j _ y (from no assumptions), then there is a proof of
j or there is a proof of y.
We lose this property when we add RAA.
Disjunction Property
If we drop RAA from our proof system, then the resulting proof system has the
DISJUNCTION PROPERTY:
§
if there is a proof of j _ y (from no assumptions), then there is a proof of
j or there is a proof of y.
We lose this property when we add RAA. As we’ve seen, p _ p is provable.
Disjunction Property
If we drop RAA from our proof system, then the resulting proof system has the
DISJUNCTION PROPERTY:
§
if there is a proof of j _ y (from no assumptions), then there is a proof of
j or there is a proof of y.
We lose this property when we add RAA. As we’ve seen, p _ p is provable.
But neither p nor p is provable.
Disjunction Property
If we drop RAA from our proof system, then the resulting proof system has the
DISJUNCTION PROPERTY:
§
if there is a proof of j _ y (from no assumptions), then there is a proof of
j or there is a proof of y.
We lose this property when we add RAA. As we’ve seen, p _ p is provable.
But neither p nor p is provable.
How do we know that?
Disjunction Property
If we drop RAA from our proof system, then the resulting proof system has the
DISJUNCTION PROPERTY:
§
if there is a proof of j _ y (from no assumptions), then there is a proof of
j or there is a proof of y.
We lose this property when we add RAA. As we’ve seen, p _ p is provable.
But neither p nor p is provable.
How do we know that?
Because our system is SOUND in the sense that if there is a proof of j, then j
is semantically valid;
Disjunction Property
If we drop RAA from our proof system, then the resulting proof system has the
DISJUNCTION PROPERTY:
§
if there is a proof of j _ y (from no assumptions), then there is a proof of
j or there is a proof of y.
We lose this property when we add RAA. As we’ve seen, p _ p is provable.
But neither p nor p is provable.
How do we know that?
Because our system is SOUND in the sense that if there is a proof of j, then j
is semantically valid; and neither p nor p is semantically valid (true under all
valuations), so it follows that neither p nor p is provable.
Next Up
Next, we will conclude our series on natural deduction for propositional logic with
an important rule: _ Elimination or “proof by cases.”
Natural Deduction XI:
Disjunction Elimination
PHIL 12A – Introduction to Logic
with Wes Holliday
University of California, Berkeley
Using Disjunctions
Suppose you’ve established a disjunction a _ b at some stage of a proof.
Using Disjunctions
Suppose you’ve established a disjunction a _ b at some stage of a proof.
How can you use this to prove some further j?
Using Disjunctions
Suppose you’ve established a disjunction a _ b at some stage of a proof.
How can you use this to prove some further j?
Show that if a holds, then j holds, and also if b holds, then again j holds.
Using Disjunctions
Suppose you’ve established a disjunction a _ b at some stage of a proof.
How can you use this to prove some further j?
Show that if a holds, then j holds, and also if b holds, then again j holds.
Since in either case, j holds, you can conclude that j holds.
Using Disjunctions
Suppose you’ve established a disjunction a _ b at some stage of a proof.
How can you use this to prove some further j?
Show that if a holds, then j holds, and also if b holds, then again j holds.
Since in either case, j holds, you can conclude that j holds.
This important method of proof is called “proof by cases.”
Using and Proving Disjunctions
Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is
divided by 4 is 0 or 1.
Using and Proving Disjunctions
Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is
divided by 4 is 0 or 1.
Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd.
Using and Proving Disjunctions
Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is
divided by 4 is 0 or 1.
Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd.
Case 1: n is even.
Using and Proving Disjunctions
Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is
divided by 4 is 0 or 1.
Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd.
Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 ,
Using and Proving Disjunctions
Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is
divided by 4 is 0 or 1.
Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd.
Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the
remainder when 4k 2 is divided by 4 is 0.
Using and Proving Disjunctions
Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is
divided by 4 is 0 or 1.
Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd.
Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the
remainder when 4k 2 is divided by 4 is 0.
Case 2: n is odd.
Using and Proving Disjunctions
Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is
divided by 4 is 0 or 1.
Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd.
Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the
remainder when 4k 2 is divided by 4 is 0.
Case 2: n is odd. Then n = 2k + 1 for some k, so n2 = (2k + 1)2 = 4k 2 + 4k + 1,
Using and Proving Disjunctions
Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is
divided by 4 is 0 or 1.
Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd.
Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the
remainder when 4k 2 is divided by 4 is 0.
Case 2: n is odd. Then n = 2k + 1 for some k, so n2 = (2k + 1)2 = 4k 2 + 4k + 1,
and the remainder when 4k 2 + 4k + 1 is divided by 4 is 1.
Using and Proving Disjunctions
Lemma 2. Let n be a nonnegative integer. Then the remainder when n2 is
divided by 4 is 0 or 1.
Proof. By Lemma 1 (which we’ll finish proving in a minute), n is even or n is odd.
Case 1: n is even. Then n = 2k for some k, so n2 = (2k )2 = 4k 2 , and the
remainder when 4k 2 is divided by 4 is 0.
Case 2: n is odd. Then n = 2k + 1 for some k, so n2 = (2k + 1)2 = 4k 2 + 4k + 1,
and the remainder when 4k 2 + 4k + 1 is divided by 4 is 1.
Since in either case, the remainder when n2 is divided by 4 is 0 or 1, we are done.
We can represent the preceding proof as follows:
1
n is even _ n is odd
2
..
.
n is even
..
.
i
remainder of n2 /4 is 0
i +1
remainder of n2 /4 is 0 _ remainder of n2 /4 is 1
j
..
.
n is odd
..
.
k
remainder of n2 /4 is 1
k +1
remainder of n2 /4 is 0 _ remainder of n2 /4 is 1
k +2
remainder of n2 /4 is 0 _ remainder of n2 /4 is 1
_I, i
_I, k
_E, 1, 2–i + 1, j–k + 1
_ Elimination
_ Elimination states that if you have a correct partial proof containing a _ b,
followed by a subproof of j from a, followed by a subproof of j from b, then you
can correctly add j on the next line in the same column as a _ b:
i
a_b
i +1
..
.
a
..
.
j
j
j +1
..
.
b
..
.
k
j
n
j
_E, i, i + 1–j, j + 1–k
Using and Proving Disjunctions
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Using and Proving Disjunctions
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
Using and Proving Disjunctions
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
Using and Proving Disjunctions
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
For the inductive step, suppose that n is even or odd.
Using and Proving Disjunctions
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
For the inductive step, suppose that n is even or odd.
Case 1: n is even.
Using and Proving Disjunctions
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
For the inductive step, suppose that n is even or odd.
Case 1: n is even. Then n = 2k for some k, so n + 1 = 2k + 1 and hence n + 1
is odd, so n + 1 has the property.
Using and Proving Disjunctions
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
For the inductive step, suppose that n is even or odd.
Case 1: n is even. Then n = 2k for some k, so n + 1 = 2k + 1 and hence n + 1
is odd, so n + 1 has the property.
Case 2: n is odd.
Using and Proving Disjunctions
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
For the inductive step, suppose that n is even or odd.
Case 1: n is even. Then n = 2k for some k, so n + 1 = 2k + 1 and hence n + 1
is odd, so n + 1 has the property.
Case 2: n is odd. Then n = 2k + 1 for some k, so n + 1 = 2k + 2 = 2(k + 1)
and hence n + 1 is even, so n + 1 has the property.
Using and Proving Disjunctions
Lemma 1. Let n be a nonnegative integer. Then n is even or odd.
Proof. We prove by induction that every n has the property that it is even or odd.
For the base case of n = 0, since 0 = 2 ˆ 0, 0 is even, so 0 has the property.
For the inductive step, suppose that n is even or odd.
Case 1: n is even. Then n = 2k for some k, so n + 1 = 2k + 1 and hence n + 1
is odd, so n + 1 has the property.
Case 2: n is odd. Then n = 2k + 1 for some k, so n + 1 = 2k + 2 = 2(k + 1)
and hence n + 1 is even, so n + 1 has the property.
We conclude that n + 1 has the property, which completes the proof.
We can represent the inductive step of the preceding proof as follows:
1
n is even _ n is odd
2
..
.
n is even
..
.
i
n + 1 is odd
i +1
n + 1 is even _ n + 1 is odd
j
..
.
n is odd
..
.
k
n + 1 is even
k +1
n + 1 is even _ n + 1 is odd
k +2
k +3
n + 1 is even _ n + 1 is odd
(n is even _ n is odd) Ñ (n + 1 is even _ n + 1 is odd)
_I, i
_I, k
_E, 1, 2–i + 1, j–k + 1
ÑI, 1–k + 2
Ex.
1
n
pÑq
p_q
Ex.
1
..
.
..
.
pÑq
2
p_ p
n
p_q
Ex.
1
..
.
2
..
.
pÑq
p_ p
3
..
.
p
..
.
i
p_q
i +1
..
.
..
.
n´1
p_q
n
p
p_q
_E, 2, 3–i, i + 1–n ´ 1
Ex.
1
..
.
2
..
.
pÑq
p_ p
3
p
4
p_q
5
..
.
p
..
.
n´1
p_q
n
p_q
_I, 3
_E, 2, 3–4, 5–n ´ 1
Ex.
1
..
.
2
..
.
pÑq
p_ p
3
p
4
p_q
5
p
6
..
.
..
.
pÑq
n´1
p_q
n
p_q
_I, 3
R, 1
_E, 2, 3–4, 5–n ´ 1
Ex.
1
..
.
2
..
.
pÑq
p_ p
3
p
4
p_q
5
p
6
pÑq
_I, 3
R, 1
7
q
ÑE, 5–6
8
p_q
_I, 7
9
p_q
_E, 2, 3–4, 5–8
Putting it All Together
We now have a formal proof system for our full language of proposition logic:
§
§
§
§
§
§
§
Reiteration;
Ñ Introduction, Ñ Elimination;
^ Introduction, ^ Elimination;
Ø Introduction, Ø Introduction;
Introduction, Elimination;
Reductio Ad Absurdum;
_ Introduction, _ Elimination.
Putting it All Together
We now have a formal proof s …
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