Trying to understand implication I'm currently slogging through propositional calculus and making my brain do impressions of a pretzel, but I'm slowly getting it though I'd like to see if that's actually true for the problem below. Is my answer correct for the below proposition?:
(p $\rightarrow$ p) $\lor$ (q $\rightarrow$ q) $\rightarrow$ p $\lor$ q
My p and q truth table is oriented with 0 and 0, and at the bottom is 1 and 1. My final result came to:

(p $\rightarrow$ p) $\lor$ (q $\rightarrow$ q) $\rightarrow$ p $\lor$ q
0
1
1
1

Sorry for the awful formatting, but I'm having trouble trying to whip up a truth table on this. Hopefully you get the idea? In
 A: Your table for $$[(p \implies p) \lor (q \implies q)]\implies (p \lor q)$$
is equivalent to proposition: $\;p\lor q$, which has precisely the same truth-table assignment as does your statement.
$\qquad\qquad\qquad\qquad\qquad$ 
Both of the Or'ed statements $(p \implies p),\;(q \implies q)$ are tautologies (true regardless of the truth values of p, and of q), and so also is, the disjunction $$(p \implies p) \lor (q \implies q)$$ It is true no matter what the truth values of $p$ and $q$. So the only way your statement can be false is if the consequence of the main implication is false: the consequence of the main implication is the disjunction $(p\lor q).\;$ And $\;p \lor q\;$ is false if and only if  $p$ is false and $q$ is false.
A: The left side is always true. So the whole thing is true except when $p$ and $q$ are both false. If your truth table says that, fine. 
For $p$ true, $q$ true, result is true.
For $p$ true, $q$ false, result is true.
For $p$ false, $q$ true, result is true.
For $p$ false, $q$ false, result is false.
Thus the sentence is equivalent ot $p\lor q$.
A: A different perspective, less helpful in context:
$p \implies p$ is true by the law of identity.
Thus $(p\implies p)\lor(q\implies q)$ is true by the law of addition.
Thus the proposition you are considering can be simplified to $\top \implies p \lor q$.
If $\top\implies p\lor q$, then $p\lor q$ is true by modus ponens.
If $p \lor q$ is true, then since every proposition implies a true statement, $\top \implies p\lor q$.
Thus your proposition is equivalent to $p \lor q$.
