Why if we assume p and ¬p we can prove any statememt?


Every proof I find is a truth table, could somebody explain it to me intuitively?

In: 1

This is easily shown in classical bivalent logic.

1. p AND NOTp
2. p (1)
3. p OR q (2, ADD)
4. NOTp (1)
5. q (3,4)

You’ve just shown any statement (q) can be derived from a contradiction.

very intuitively: Think of proof by contradiction. assuming p and !p lets you do proofs by contradiction on demand. “Assume unicorns are not real. We know p. we know !p. This is a contradiction. Therefore unicorns are real.”

A bit more formally:

assume the statement (p AND !p) is always true. Therefore, the statement (q OR (p AND !p)) is also always true for any q. Then by boolean algebra rules, ((q OR p) AND (q or !p)) is true.

so, either one of q and p are true, and either one of q and !p are true.

but if we assume that p is true, then !p in false. But (q OR !p) is true. since either q or !p is true, and !p is false, q must be true.

Let’s say unicorns exist. And let’s say unicorns don’t exist.

Now take the statement “Either unicorns exist or Fjortoftsairplane is the greatest of all time”.

We know that statement is true because “Unicorns exist” is true.

“Unicorns exist” is false. Which means the bit after the “or” must be true.

Congratulations, we just proved I am the greatest of all time.

Because if you had both P is true and P is false, you would have no way to prove anything else (such as Q) as false, no matter what the relation is between P and Q .