Value V on some formula in Logic I want to calculate, how many value $v$ on {$p,q,r$} has, such that sentence $(p \to (q\wedge r)) \to r$ gets value $0$?
I solve it via truth table, any other methods for solving such questions? or using  truth tables in disguise?
 A: You'll always have to use "truth tables in disguise", but you can try to make the process more efficient. For one, instead of deriving the full truth tables for every sentence letters, you can try to minimize the number of operations you need to do to reach the solution.
You are only interested in the case(s) where your proposition returns $0$ (or "False"). You could make the following steps:
1) When is $(p\to(q\land r))\to r$ false? When the premise is true and the conclusion false: therefore $v(r)=0$.
2) When is $p\to(q\land r)$ true? There are three cases, but knowing that $v(r)=0$ automatically makes the conclusion false, such that you are down to only one case, namely when the premise and the conclusion are both false. Therefore $v(p)=0$ and there is no restriction on $v(q)$.
Without working out the full truth tables (8 cases), you can find out in two simple steps that there are two assignments of truth values that make the proposition false: $v(p,q,r)=(0,0,0)$ and $v(p,q,r)=(0,1,0)$.
A: $(p\to(q\wedge r))\to r$ can be simplified via
$
 \Updownarrow  (\text{implication equivalence, twice})\\
\neg(\neg p\vee (q\wedge r))\vee r
\\ \Updownarrow (\text{de Morgan's negation, twice}) \\
(p\wedge (\neg q\vee\neg r))\vee r
\\ \Updownarrow (\text{distribution}) \\
(p\vee r)\wedge ((\neg q\vee \neg r)\vee r)
\\ \Updownarrow (\text{tautology, via associativity}) \\
(p\vee r)\wedge (\neg q\vee \top)
\\ \Updownarrow (\text{absorption}) \\
(p\vee r)\wedge \top
\\ \Updownarrow (\text{identity}) \\
p\vee r$ 
So the number of distinct triplets $(p, q, r)$ that make $p\vee r$ true are $2\times 3$ .  Thus to make it false you need....?
