[ negation p and (p or q)] implies q.
I have no idea how to start.
Edit: Maybe it will make more intuitive sense if you put statements in.
"It is true that I'm not happy and that: I'm happy or rich (or both)." Do you see why this in all cases implies that "I'm rich"?
Edit 2: Sorry didn't notice the "without truth tables" part.
This might not be quite right, but here's my attempt
not p and (p or q) <-> (not p and p) or (not p and q) <-> not p and q
so we have
which is a tautology because the conclusion is the same as one of the premises.
Just for fun, I'm going to prove this formula from a "programmer's view".Prove without truth tables: .
In the programmer's view, propositions like , , etc. are types like for natural numbers, for Booleans and for strings of characters. Next, is a function that accepts and returns ; is a pair of and , and is a disjoint union of and .
I will write for "false". Then is equivalent to . Also, is a tautology for any . Thus, is a type such that, given , one can produce a value of any type.
Now, towards the proof of the formula, assume we are given . This is a pair of, first, and, second, either or . We need to produce . If the second given element is , then we are done. If it is , then, using the first element , we can manufacture any type, including .