Proof of logical equivalence I have been trying to get my head around this but currently online classes are horrible plus there is not many instructions but how do i get from $(P \wedge Q) -> (P \vee Q)$ to $(P>Q)$?
Based on what I understood we use the logical conditional statement
$P \to Q =$ $\sim P \vee Q$
to get $\sim (P \vee Q) \vee (P \vee Q)$.
Then we use De Morgan's Law to get $(\sim P$ $\vee \sim Q) \vee (P \vee Q)$.
After we use the associate and communicative law we end up with $(\sim P$ $\vee \sim Q) \vee (P \vee Q)$ which is a logical equivalence to $(P \to Q)$ because $P \to Q = \sim P \vee Q$.
Could you please correct me if i got it wrong, sorry for the format but its my first post and i dont know how to fix it yet.
 A: There's something wrong, because $P\wedge Q\rightarrow P\vee Q$ is always true. You can see this by doing the calculations as you've done them, thus getting $$(\sim P\vee\sim Q)\vee(P\vee Q)=(\sim P\vee P)\vee(\sim Q\vee Q)=\text{true}\vee\text{true}=\text{true}$$
or, just by looking at the meaning of the implication, which is "If $P$ and $Q$ are true, then $P$ or $Q$ are true", which is clearly true. In particular, it is true even if $P\rightarrow Q$ is false, which means that they cannot be equivalent.
A: 
$\sim (P \vee Q) \vee (P \vee Q)$.
Then we use De Morgan's Law to get $(\sim P \vee \sim Q) \vee (P \vee Q).$

You made a mistake here: the second line ought to instead be $$(\sim P \land \sim Q) \vee (P \vee Q).$$

But how do i get from  $$(P \wedge Q) \to (P \vee Q)$$ to $$(P\to Q)?$$

I suspect that you've copied the given question wrongly. As pointed out by Alessandro, the former is a tautology. So, you're basically being asked to prove that the latter is a tautological consequence of a tautology. But this is the same asking you to prove that the latter is a tautology. This is impossible, because the latter is patently not a tautology (i.e., always true no matter what the truth values of $P$ and $Q$ are).
