Q.Construct the truth table for: $[(\sim p \vee q) \wedge (q \to r)] \to (p \to r)$

Build the 8-row table over $p,q,r$. Let $A=\sim p\vee q$ (equivalently $p\to q$), $B=q\to r$, $C=p\to r$; the pattern is $(A\wedge B)\to C$.

$p{=}T,q{=}T,r{=}T$: $A=T$, $B=T$, $A\wedge B=T$; $C=T$; result$=T\to T=T$.

$p{=}T,q{=}T,r{=}F$: $A=T$, $B=F$, $A\wedge B=F$; $C=F$; result$=F\to F=T$.

$p{=}T,q{=}F,r{=}T$: $A=F$, $B=T$, $A\wedge B=F$; $C=T$; result$=F\to T=T$.

$p{=}T,q{=}F,r{=}F$: $A=F$, $B=T$, $A\wedge B=F$; $C=F$; result$=F\to F=T$.

$p{=}F,q{=}T,r{=}T$: $A=T$, $B=T$, $A\wedge B=T$; $C=T$; result$=T\to T=T$.

$p{=}F,q{=}T,r{=}F$: $A=T$, $B=F$, $A\wedge B=F$; $C=T$; result$=F\to T=T$.

$p{=}F,q{=}F,r{=}T$: $A=T$, $B=T$, $A\wedge B=T$; $C=T$; result$=T\to T=T$. …

Compute helper columns $A=\sim p\vee q$, $B=q\to r$, $C=p\to r$, then combine as $ …

Treating $\sim p\vee q$ as unrelated to $p\to q$ instead of recognising it as the same thing, which makes the p …