So, we want to show that p is true.
In a proof by contradiction, we start out by saying:
We then show that this leads to a contradiction, a statement like (q ∧ ¬q)
That is, we show that ¬p→(q ∧ ¬q), where (q ∧ ¬q) is some contradiction
But, how does this allow us to conclude that p is true?
Well, we know that (q ∧ ¬q) is always false. Thus if we show that ¬p→(q ∧ ¬q), we have shown that ¬p→F.
That is, we have shown that (¬p→F) is a true statement.
And we know that a→b is equivalent to writing ¬a ∨ b
| a | b | a→b | ¬a | b | ¬a ∨ b |
|---|---|---|---|---|---|
| T | T | T | F | T | T |
| T | F | F | F | F | F |
| F | T | T | T | T | T |
| F | F | T | T | F | T |
Thus if (¬p→F) is true, then (¬(¬p ) ∨ F) is true, i.e. ( p ∨ F) is true
By the identity law ( p ∨ F) ≡ p we have that p is true.
By now, you should see that these tables on p. 24 and 25 are pretty handy. If you haven't read through the textbook sections that surround these tables, now would be a good time to do that.
P. Conrad, 04/09/2008