# When is $(p_1 \rightarrow (p_3 \rightarrow (\lnot p_4 \rightarrow p_2)))$ false? [closed]

I found that the statement:

$$(p_1 \rightarrow (p_3 \rightarrow (\lnot p_4 \rightarrow p_2)))$$

can only be false when $p_1 = T = p_3$ and $p_2 = F = p_4$. Is this right? Sorry but I can't just draw the truth table which consists of $16$ rows to show my work

Let's cover $(p1 \rightarrow (p3 \rightarrow (\lnot p4 \rightarrow p2)))$ one element at a time
$p1 \rightarrow(p3 \rightarrow(\lnot p4 \rightarrow p2))$ can only be false if p1 is true and $p3 \rightarrow(\lnot p4 \rightarrow p2)$ is false
So $p1$ is true, onwards $p3 \rightarrow(\lnot p4 \rightarrow p2)$
$p3$ is true and $\lnot p4 \rightarrow p2$ is false
$\lnot p4 \rightarrow p2$
So $\lnot p4$ is true and \rightarrow p2$$is false So p4 is false p1 = p3 = true p2 = p4 = false Yes you're right • Very simple! Maybe this would have occurred to me had I not been practicing truth tables (I'll accept the answer as soon as I can) – user290300 Commented Aug 18, 2016 at 13:57 It might help you to better understand "why" the required truth-values must be assigned in order to make the statement true (and hence its negation false). Perhaps you've encountered the logical equivalence that follows:$$p\rightarrow (q\rightarrow r) \equiv (p\land q) \rightarrow r$$(If you haven't encountered it yet, prove it using a truth-table to confirm.) Now, given$$\begin{align}(p_1 \rightarrow (p_3 \rightarrow (\lnot p_4 \rightarrow p_2))) \\\\ &\equiv p_1 \rightarrow ((p_3 \land \lnot p_4) \rightarrow p_2)\\\\ &\equiv (p_1 \land p_3 \land \lnot p_4) \rightarrow p_2\\\\ &\equiv \lnot[p_1\land p_3 \land \lnot p_4] \lor p_2\\\\ &\equiv \lnot [p_1 \land p_3 \land \lnot p_4 \land \lnot p_2] \end{align}$$I twice used the equivalence I referred to above, then I used the equivalence (a\rightarrow b) \equiv ((\lnot a) \lor b). Finally, I use one of DeMorgan's rules. (Also used implicitly is the fact that conjunction is associative.) The truth of the original statement is equivalent to the final equivalent expression. To make each statement false, we simply negate it, leaving$$\lnot\{\lnot [p_1 \land p_3 \land \lnot p_4 \land \lnot p_2]\}\equiv [p_1 \land p_3 \land \lnot p_4 \land \lnot p_2]$$Falsehood here requires p_1 is true, AND p_3 is true, AND (\lnot p_4) is true (hence p_4 is false), AND \lnot p_2 is true (hence p_2 is false.) It is true.. In order for implication, say A \rightarrow B, to be false, A needs to be true and B false. That means that p_1 needs to be true, and the rest of the statement false. Similar reasoning says p_3 needs to be true and the rest of the statement false. Again, p_4 is false cause \lnot p_4 must be true and p_2 is false..$$\begin{array}{rl} p_1 \to (p_3 \to (\neg p_4 \to p_2)) &\equiv p_1 \to (p_3 \to (\neg \neg p_4 \lor p_2))\\ &\equiv p_1 \to (p_3 \to (p_4 \lor p_2))\\ &\equiv p_1 \to (\neg p_3 \lor (p_4 \lor p_2))\\ &\equiv \neg p_1 \lor (\neg p_3 \lor (p_4 \lor p_2))\\ &\equiv \neg p_1 \lor p_2 \lor \neg p_3 \lor p_4\end{array}$which is false only when$p_1 = p_3 = \text{True}$and$p_2 = p_4 = \text{False}\$.