Solving Logical equivalence & propositional logic problems without truth tables I have no particular "Logic question" in hand at the time being, but need help to understand a way that can be used to prove "Logical equivalence without using truth tables".
moreover can we solve all propositional logic questions without truth tables. explain with some basic examples please (there are tons of examples for this topic but i found those very complex when in it comes to understanding). 
explanation a solving this without a truth table i think will do for me.
p  ∧ (q ∨ r) = (p ∧ q) ∨ (p ∧ r) 
 A: Yes, for example you can use DeMorgan's laws:
$$a \vee b = \neg((\neg a) \wedge (\neg b))$$
$$a \wedge b = \neg((\neg a) \vee (\neg b))$$
As well as distributivity over the operators:
$$a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c) $$
$$a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c) $$
And there are a lot of more rules you can use to simplify expressions. 
A: As long as you are using a sound deductive system, to prove $\phi\equiv\psi$ you have to assume $\phi$ then deduce $\psi$ from this assumption and assume $\psi$ then deduce $\phi$ from this assumption. Since a sound deductive system is truth preserving, this ensures that if $\phi$ is true then $\psi$ is true and if $\psi$ is true then $\phi$ will be true - that is, that $\phi$ and $\psi$ have the same truth values. 
I'm not sure what you consider a simple example because I'm not sure what deductive system you are familiar with. I suggest considering $\neg\neg B\equiv B$. Pick your sound deductive system of choice. If you can deduce $B$ from the assumption $\neg\neg B$ and also deduce $\neg\neg B$ from the assumption of $B$, then this will prove the equivalence without a truth table, since a sound deduction can never lead to a false conclusion if the assumption is true. 
Another option is to use a semantic proof, which requires you to know the semantic truth conditions for logical formula. Here is an example from propositional logic to prove $\neg\neg B\equiv B$. Again, you have to assume each proposition is true and show via semantic rules that the other proposition is true. Let $v$ be a valuation such that $v(\neg\neg B)=true$. Then $v(\neg B)=false$, which implies that $v(B)=true$. In the other direction, let $v'$ be a valuation such that $v'(B)=true$. Then $v'(\neg B)=false$, which implies that $v'(\neg\neg B)=true$. Since $v$ and $v'$ are arbitrary, for any valuation $v$, $v(\neg\neg B)=v(B)$. Therefore, $\neg\neg B\equiv B$. 
