If the proposition $\lnot p \to q$ is true, then the truth value of the proposition $\lnot p \lor \left ( p \to q \right )$, where $\lnot$ is negation, $\lor$ is inclusive OR and $\to$ is implication, is
Cannot be determined.
From the axiom $\lnot p \to q$, we can conclude that $p \vee q$.
So, either $p$ or $q$ must be TRUE.
&\lnot p \lor (p \to q)\\[1em]
\equiv& \lnot p \lor ( \lnot p \lor q)\\[1em]
\equiv&\lnot p \lor q
Since nothing can be said about the Truth value of $p$, it implies that $\lnot p \lor q$ can also be True or False.
Hence, the value cannot be determined.
@Pragy Agarwal Thank you for correcting me. I understood the mistake I made.
given that $ \lnot p \to q$ is true
A proposition can have only 2 possible values namely true or false. Here I feel “cannot be determined”(option D) is a more suitable answer than “multiple values”(option B) because propositional variables are similar to variables in programming which can have only one value at a time.
We need to look only for the GREEN case as in that only P’ → Q is true .
And for these three (GREEN) cases we have truth value for some cases and false for some cases .
So the value can’t be determined .
Answer will be D.