A formula is valid if it is true for any values of its variables.
Conversely, a formula is not valid, if there exists at least one combination of values of variables for which it is false.
We can use either method to solve for validity.
Case 1: Trying to prove tautology.
$[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow r’$
$=[(p’ \lor q) \land (q’ \lor r)] \rightarrow r’$
$=[(p’ \lor q) \land (q’ \lor r)]’ \lor r’$
$=(p \land q’) \lor (q \land r’) \lor r’$
Re-writing the connectives using +, *, ‘ for simplicity sake:
$=pq’ + qr’ + r’$
$=pq’ + r’$
which is NOT a tautology.
Hence, the formula is NOT valid.
Case 2: Trying to find atleast one combination where the formula yields false.
An implication $A \rightarrow B$ is false, iff $A=True$ and $B=False$
$[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow r’$
Let’s assume, $RHS = False$
i.e., $r’ = False \Rightarrow r = True$
$LHS:$
$[(p \rightarrow q) \land (q \rightarrow r)]$
$=(p’ + q) (q’ + r)$
$=p’q’ + p’r + qr$
Putting the value of $r=True \text{, or } 1$
$=p’q’ + p’.1 + q.1$
$=p’q’ + p’ + q$
$=p’ + q$
$=p \rightarrow q$
which is false, if $p=True$ and $q=False$
Since, when $RHS$ is false, $LHS$ can also become false, the original formula is NOT valid.