In parallel with this thread, I’d like to present what I offered late in this thread. That is, I’m going to show why proving both a statement and its contradiction is a Very Bad Thing.
There’s some terminology and notation that we need to get out of the way first. A statement is anything that can be either true or false. Some statements are “I like beans”, “It is raining outside”, and “Proving both a statement and its contradiction is a Very Bad Thing”. The negation of a statement A is a statement not-A which denies whatever A says. The negations of the above statements are “I don’t like beans”, “It is not raining outside”, and “Proving both a statement and its contradiction is not a Very Bad Thing”. Obviously, A and not-A can’t both be true at the same time; it should be obvious that not-A is true exactly when A is false.
Suppose we have two statements, A and B. Let’s define two new operations, “and” and “or”. A-and-B is true only when A is true and B is also true. A-or-B is only false when A is false and B is also false.
If you notice a strong similarity between these two definitions, you get a gold star; one can easily show that not-(A-and-B) is true exactly when (not-A)-or-(not-B) is true. Parentheses denote operations to be performed first, much like they do in ordinary arithmetic.
This isn’t standard notation, and it’s taking a bit out of me to write it this way. So: ~A is short for not-A, A v B is short for A-or-B, and A & B is short for A-and-B. So that I don’t have to explain operator precedence, I will use parentheses whenever there’s any ambiguity as to the order in which I mean operations to be performed.
In case you didn’t guess, a contradiction is a statement of the form A & (~A). Is it obvious that this can’t be true, or should I explain it further?
All right, the preliminaries are over, so here we go. These next few paragraphs are kinda heady, so strap yourself in.
For the sake of argument, let’s say that we have proved a contradiction, A & (~A). From the definition of “&” above, we know that A & (~A) can only be true if A is true and (~A) is true. So we know that A is true.
Now what about A v B? This could only be false if A is false and B is false. But we already know that A is true, so A v B must be true.
Well, if A v B is true, it follows that at least one of A and B is true. We know from our original contradiction that ~A is true, so A must be false. If A is false, A v B can only be true if B is true. Therefore, it follows that B is true.
This explains why it would be bad to prove both a statement and it’s negation. Clearly, some statements must be false; however, if a contradiction is true, every statement is automatically true.
I know I’m gonna get accused of being disingenuous, but I’ll swear on anything you want that this argument really does hold water. So fire up those questions, and let’s go.