Doubt in logic theory I  am a beginner in mathematical logic. I found following which i did not understand
1) Addition logical implicative -
$p \to (p \lor q)$
In explanation, he says if we know p, then we can add q. Then he says, we can know about p or q. I did not understand the intuition behind this. If p is true then when we add q(irrespective of true or false), we can say p or q is true. But when p is false, we can not know about p or q right?
Pls note that i can prove that the above is tautology
2)Simplification -
$(p \land q) \to p$ - in explanation he says if we know p and q then we can know about p. How? if p and q is true then we can say that p was true. But then if p and q is false, we can not say about p right?. Pls note that i can prove that the above is tautology
3) Distributive law wrto implication
I understood distributive law wrto and and or operators.
$p \to (q \land r) \iff (p \to q) \land (p \to r)$
Do we have similar distributive law wrto following implication.
Say $[p \lor (q \implies r)] \iff [(p \lor q) \implies (p \lor r)]$
 A: In parts (1) and (2) You seem to be confusing material implication $A \rightarrow B$ with material equivalence $A \leftrightarrow B$.
$A \rightarrow B$ is true as long as it is not the case that $A$ is true and $B$ is false. In symbols:
$(A \rightarrow B) \leftrightarrow \lnot(A \land \lnot B)$
So if we know that $A$ is true and $A \rightarrow B$ then we can conclude that $B$ is true. But if we know that $A$ is false then $A \rightarrow B$ tells us nothing; $B$ can be either true or false.
On the other hand, $A \leftrightarrow B$ means that $A$ and $B$ have the same truth value.
For part (3) use a truth table. List the eight combinations of truth values for $p, q, r$ and evaluate $p \lor (q \rightarrow r)$ and $(p \lor q) \rightarrow (p \lor r)$ for each combination. If $p \lor (q \rightarrow r)$ and $(p \lor q) \rightarrow (p \lor r)$ always have the same truth value then $(p \lor (q \rightarrow r)) \leftrightarrow ((p \lor q) \rightarrow (p \lor r))$. If not then you have found a counterexample.
