Question 1193814: Prove using Indirect Proof
1. B ⊃ (K • M)
2. (B • M) ⊃ (P ≡ ∼P) / ∼B
Found 3 solutions by Alan3354, math_tutor2020, ikleyn:
Answer by Alan3354(69443)   (Show Source):
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
This is one way to do the derivation using an indirect proof (aka proof by contradiction).
| Number | Statement | Line(s) Used | Reason | | 1 | | B -> (K & M) | | | | 2 | | (B & M) -> (P = ~P) | | | | :. | | ~B | | | | 3 | B | | Assumption for Indirect Proof | | 4 | K & M | 1,3 | Modus Ponens | | 5 | M | 4 | Simplification | | 6 | B & M | 3, 5 | Conjunction | | 7 | P = ~P | 2, 6 | Modus Ponens | | 8 | (P -> ~P) & (~P -> P) | 7 | Material Equivalence | | 9 | P -> ~P | 8 | Simplification | | 10 | ~P v ~P | 9 | Material Implication | | 11 | ~P | 10 | Taulogy | | 12 | ~P -> P | 8 | Simplification | | 13 | ~~P v P | 12 | Material Implication | | 14 | P v P | 13 | Double Negation | | 15 | P | 14 | Taulogy | | 16 | P & ~P | 15, 11 | Conjunction | | 17 | | ~B | 3 - 16 | Indirect Proof |
The conclusion we want to arrive at is ~B
Assume that the opposite is the case and we assume B (line 3)
The idea is to see if it generates a contradiction.
The line P & ~P is a contradiction because one portion is true while the other is false, which makes P & ~P always false.
Or perhaps another approach is to look at line 7 where we have P = ~P. This isn't possible because if P is true then ~P is false, and vice versa. There's no way to have true equal false. So we could have stopped at that point to find the contradiction.
Therefore, the negation of the assumption must be the case and we have ~B instead as the proper conclusion.
I used arrows in place of horseshoe symbols.
I used ampersands (&) in place of the dots.
Instead of a triple equal sign, I used a regular equal sign.
Answer by ikleyn(52777)  (Show Source):
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