- #1

member 587159

I want to proof that:

##p \land (p \to q) \Rightarrow q##

I know this is a quite trivial problem using truth tables, however, I want to do it without it. As I'm learning this myself, is this the correct approach?

##p \land (p \to q)##

##\iff p\land (\neg p \lor q)##

##\iff (p \land \neg p) \lor (p \land q)## (distributive law)

Now, ##p \land \neg p## is a contradiction (see the questions below)

##\iff (p \land q)##

Now, it is clear, that ##p \land q \Rightarrow q##

Also, I have some additional questions.

1) Suppose there is a contradiction (or tautology) r. Can we always say then, that:

##r \lor p \iff p##

2) Suppose there is a contradiction r. Can we always say then, that:

##r \land p## is a contradiction?

3) Suppose there is a tautology r. Can we always say then, that:

##r \land p \iff p##

Thanks in advance.