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


4 Answers 4


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

  • $\begingroup$ 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) $\endgroup$
    – 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 $p_2 = p_4 = \text{False}$.