If p,q and r are three propositions, then which of the following combination of truth values of p, q and r makes the logical expression…

Question

If p,q and r are three propositions, then which of the following combination of truth values of p, q and r makes the logical expression {(p∨q)∧((~p)∨r)}→((~q)∨r) false ?

MARKS Content Team · Accepted Answer

Correct answer: (C) p=F,q=T,r=F. Given, {(p∨q)∧((~p)∨r)}→((~q)∨r) Now solving the above expression using truth table from given options we get, p q r p∨q ~p∨r A≡p∨q∧~p∨r B≡~q∨r A→B A T F T T T T T T B T T F T F F F T C F T F T T T F F D T F F T F F T T So, only Option C will be false.

	$p$	$q$	$r$	$(p \lor q)$	$(~ p \lor r)$	$A \equiv (p \lor q) \land (~ p \lor r)$	$B \equiv ~ q \lor r$	$A \to B$
$(A)$	T	F	T	T	T	T	T	T
$(B)$	T	T	F	T	F	F	F	T
$(C)$	F	T	F	T	T	T	F	F
$(D)$	T	F	F	T	F	F	T	T

If $p, q$ and $r$ are three propositions, then which of the following combination of truth values of $p, q$ and $r$ makes the logical expression ${(p \lor q) \land ((~ p) \lor r)} \to ((~ q) \lor r)$ false ?

Solution

Relevant Questions