Zeroth-Order Natural Deduction. Part 4: Indirect Proof
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1 Zeroth-Order Natural Deduction Part 4: Indirect Proof
2 Conditional Proof Last time, we introduced the rule of conditional proof, which lets us discharge assumptions. Today, we will begin by looking at some more examples.
3 Conditional Proof Here is our first example for today. Take two or three minutes to talk to your neighbors about how to do this proof. { (P Q) } ((P ᴧ R) Q)
4 Conditional Proof Assumptions Line Formula Justification 1 (1) (P Q) A
5 Conditional Proof Assumptions Line Formula Justification 1 (1) (P Q) A 2 (2) (P ᴧ R) A (for CP)
6 Conditional Proof Assumptions Line Formula Justification 1 (1) (P Q) A 2 (2) (P ᴧ R) A (for CP) When proving a conditional, assume its antecedent. Then derive its consequent and use the rule of conditional proof!
7 Conditional Proof Assumptions Line Formula Justification 1 (1) (P Q) A 2 (2) (P ᴧ R) A (for CP) 2 (3) P 2 ᴧE
8 Conditional Proof Assumptions Line Formula Justification 1 (1) (P Q) A 2 (2) (P ᴧ R) A (for CP) 2 (3) P 2 ᴧE 1,2 (4) Q 1,3 E
9 Conditional Proof Assumptions Line Formula Justification 1 (1) (P Q) A 2 (2) (P ᴧ R) A (for CP) 2 (3) P 2 ᴧE 1,2 (4) Q 1,3 E 1 (5) ((P ᴧ R) Q) 2,4 CP
10 Conditional Proof Assumptions Line Formula Justification 1 (1) (P Q) A 2 (2) (P ᴧ R) A (for CP) 2 (3) P 2 ᴧE 1,2 (4) Q 1,3 E 1 (5) ((P ᴧ R) Q) 2,4 CP
11 Conditional Proof A new kind of example. Given no premisses at all, we want to prove the conclusion (P P). { } (P P)
12 Conditional Proof A new kind of example. Given no premisses at all, we want to prove the conclusion (P P). { } (P P) How can we do this?
13 Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) (2) (P P) 1 CP
14 Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) (2) (P P) 1 CP
15 Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) (2) (P P) 1 CP
16 Review: Conditional Proof One final example. Talk to your neighbors about how to approach this one. { } (P ((P Q) Q))
17 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) Why assume that P? If you are trying to prove a conditional, try assuming the antecedent and using the rule of conditional proof. (P ((P Q) Q))
18 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) After assuming that P, we need to prove that ((P Q) Q), so we assume that (P Q). (P ((P Q) Q))
19 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E
20 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP
21 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP
22 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP
23 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP
24 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP
25 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP
26 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP
27 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP (5) (P ((P Q) Q)) 1,4 CP
28 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP (5) (P ((P Q) Q)) 1,4 CP
29 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP (5) (P ((P Q) Q)) 1,4 CP
30 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP (5) (P ((P Q) Q)) 1,4 CP
31 Review: Conditional Proof Assumptions Line Formula Justification 1 (1) P A (for CP) 2 (2) (P Q) A (for CP) 1,2 (3) Q 1,2 E 1 (4) ((P Q) Q) 2,3 CP (5) (P ((P Q) Q)) 1,4 CP
32 The rule of conditional proof, together with the rule for negation introduction gives us a powerful, generic proof technique: Indirect Proof
33 The rule of conditional proof, together with the rule for negation introduction gives us a powerful, generic proof technique: Indirect Proof Also called proof by contradiction or reductio ad absurdum.
34 In conditional proof, we make an assumption that is then discharged by becoming the antecedent of a material conditional. Indirect proofs are conditional proofs which assume that the conclusion we want to prove is false.
35 Sketch of how indirect proof that ~P works: Assume (for reductio) that P.
36 Sketch of how indirect proof that ~P works: Assume (for reductio) that P. Alternatively, if we want to prove that P, then: Assume (for reductio) that ~P.
37 Sketch of how indirect proof that ~P works: Assume (for reductio) that P.
38 Sketch of how indirect proof that ~P works: Assume (for reductio) that P. Use P to derive a sentence and its negation, e.g. Q and ~Q.
39 Sketch of how indirect proof that ~P works: Assume (for reductio) that P. Use P to derive a sentence and its negation, e.g. Q and ~Q. Discharge P to get a pair of conditionals like (P Q) and (P ~Q).
40 Sketch of how indirect proof that ~P works: Assume (for reductio) that P. Use P to derive a sentence and its negation, e.g. Q and ~Q. Discharge P to get a pair of conditionals like (P Q) and (P ~Q). Use negation introduction to get ~P.
41 Here is a simple example: { } ~(P ᴧ ~P)
42 Assumptions Line Formula Justification 1 (1) (P ᴧ ~P) A*
43 Assumptions Line Formula Justification 1 (1) (P ᴧ ~P) A* 1 (2) P 1 ᴧE 1 (3) ~P 1 ᴧE
44 Assumptions Line Formula Justification 1 (1) (P ᴧ ~P) A* 1 (2) P 1 ᴧE 1 (3) ~P 1 ᴧE (4) ((P ᴧ ~P) P) 1,2 CP
45 Assumptions Line Formula Justification 1 (1) (P ᴧ ~P) A* 1 (2) P 1 ᴧE 1 (3) ~P 1 ᴧE (4) ((P ᴧ ~P) P) 1,2 CP (5) ((P ᴧ ~P) ~P) 1,3 CP
46 Assumptions Line Formula Justification 1 (1) (P ᴧ ~P) A* 1 (2) P 1 ᴧE 1 (3) ~P 1 ᴧE (4) ((P ᴧ ~P) P) 1,2 CP (5) ((P ᴧ ~P) ~P) 1,3 CP (6) ~(P ᴧ ~P) 4,5 ~I
47 Alternatively, we might already accept Q: Assume (for reductio) that P.
48 Alternatively, we might already accept Q: Assume (for reductio) that P. Use P to derive a sentence, ~Q, that is in contradiction with some sentence Q that you already accept.
49 Alternatively, we might already accept Q: Assume (for reductio) that P. Use P to derive a sentence, ~Q, that is in contradiction with some sentence Q that you already accept. Discharge P to get (P ~Q), and use arrow introduction to get (P Q)
50 Alternatively, we might already accept Q: Assume (for reductio) that P. Use P to derive a sentence, ~Q, that is in contradiction with some sentence Q that you already accept. Discharge P to get (P ~Q), and use arrow introduction to get (P Q) Use negation introduction to get ~P.
51 Galileo claimed that all bodies of the same material fall at the same rate, regardless of how heavy they are. Galileo ( )
52 Galileo claimed that all bodies of the same material fall at the same rate, regardless of how heavy they are. According to legend, he tested his claim at the Leaning Tower of Pisa. Galileo ( )
53
54 The Leaning Tower of Pisa
55 Galileo
56 Hi everybody! Galileo
57
58 Look out below!
59
60
61
62
63 Take that Aristotle!
64 Although the experiment may not have actually taken place Galileo ( )
65 Galileo had an indirect proof that the experiment must work out the way we just saw. Galileo ( )
66 Galileo reasoned as follows
67 Galileo reasoned as follows. Suppose each falling body acquires a definite velocity that is fixed by nature.
68 Galileo reasoned as follows. Suppose each falling body acquires a definite velocity that is fixed by nature. Then suppose for reductio that a heavier body falls faster than a lighter body.
69 Galileo reasoned as follows. Suppose each falling body acquires a definite velocity that is fixed by nature. Then suppose for reductio that a heavier body falls faster than a lighter body. What happens if we fasten the two together?
70 If we tie them together, the heavier one will be slowed down by the lighter one, and the lighter one will be sped up by the heavier one.
71 If we tie them together, the heavier one will be slowed down by the lighter one, and the lighter one will be sped up by the heavier one. So, if we tie them together, they will acquire a velocity less than that of the heavy one.
72 However, if we tie them together, they will compose a single object heavier than either one alone.
73 However, if we tie them together, they will compose a single object heavier than either one alone. So, if we tie them together, they will acquire a velocity greater than that of the heavier one.
74 But that is absurd! The two tied together can t fall with a velocity that is both greater than and less than that of the heavier one!
75 But that is absurd! The two tied together can t fall with a velocity that is both greater than and less than that of the heavier one! Hence, we have to reject our supposition that a heavier body falls faster than a lighter body.
76 But that is absurd! The two tied together can t fall with a velocity that is both greater than and less than that of the heavier one! Hence, we have to reject our supposition that a heavier body falls faster than a lighter body. Therefore, a heavier body does not fall faster than a lighter body.
77 P = Every object reaches a natural velocity. Q = A heavier object falls faster than a lighter object. R = Two joined objects fall slower than the heavier of the two.
78 Assumptions Line Formula Justification 1 (1) P A 2 (2) ((P ᴧ Q) R) A 3 (3) ((P ᴧ Q) ~R) A
79 Assumptions Line Formula Justification 1 (1) P A 2 (2) ((P ᴧ Q) R) A 3 (3) ((P ᴧ Q) ~R) A 4 (4) Q A*
80 Assumptions Line Formula Justification 1 (1) P A 2 (2) ((P ᴧ Q) R) A 3 (3) ((P ᴧ Q) ~R) A 4 (4) Q A* 1,4 (5) (P ᴧ Q) 1,4 ᴧI
81 Assumptions Line Formula Justification 1 (1) P A 2 (2) ((P ᴧ Q) R) A 3 (3) ((P ᴧ Q) ~R) A 4 (4) Q A* 1,4 (5) (P ᴧ Q) 1,4 ᴧI 2,3 (6) ~(P ᴧ Q) 2,3 ~I
82 Assumptions Line Formula Justification 1 (1) P A 2 (2) ((P ᴧ Q) R) A 3 (3) ((P ᴧ Q) ~R) A 4 (4) Q A* 1,4 (5) (P ᴧ Q) 1,4 ᴧI 2,3 (6) ~(P ᴧ Q) 2,3 ~I 1 (7) (Q (P ᴧ Q)) 4,5 CP
83 Assumptions Line Formula Justification 1 (1) P A 2 (2) ((P ᴧ Q) R) A 3 (3) ((P ᴧ Q) ~R) A 4 (4) Q A* 1,4 (5) (P ᴧ Q) 1,4 ᴧI 2,3 (6) ~(P ᴧ Q) 2,3 ~I 1 (7) (Q (P ᴧ Q)) 4,5 CP 2,3 (8) (Q ~(P ᴧ Q)) 6 I
84 Assumptions Line Formula Justification 1 (1) P A 2 (2) ((P ᴧ Q) R) A 3 (3) ((P ᴧ Q) ~R) A 4 (4) Q A* 1,4 (5) (P ᴧ Q) 1,4 ᴧI 2,3 (6) ~(P ᴧ Q) 2,3 ~I 1 (7) (Q (P ᴧ Q)) 4,5 CP 2,3 (8) (Q ~(P ᴧ Q)) 6 I 1,2,3 (9) ~Q 7,8 ~I
85 Next Time We will begin talking about first-order logic!
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