Question

Using a truth table determine whether the argument form is valid or invalid

p ∧ q →∼ r

p∨∼q

∼q→p

∴∼ r

Answer 1

Truth Table:

p	q	r	~q	~r	(pq)	(p q) ~r	(p V `~q)	(~q p)	~q	[(p q) ~r] (p V `~q) (~q p)
0	0	0	1	1	0	1	1	0	1	0
0	0	1	1	0	0	1	1	0	1	0
0	1	0	0	1	0	1	0	1	0	0
0	1	1	0	0	0	1	0	1	0	0
1	0	0	1	1	0	1	1	1	1	1
1	0	1	1	0	0	1	1	1	1	1
1	1	0	0	1	1	1	1	1	0	1
1	1	1	0	0	1	0	1	1	0	0

Explanation:

Negation(¬) of any logical Identity P is Nothing but Opposite Truth Value for P. i.e If P is T then (¬P) is F

And() Operation is Nothing but if Truth value of P and Q is True then (PQ) is True

If any Truth value of P and Q is False then (PQ) is False

OR(V) Operation is Nothing but if Truth values of both P & Q is False then (PVQ) is False

If any Truth value of P & Q is True then (PVQ) is True

Implies() Operation is Nothing but if Truth value of (PQ) is False If and only if P has Truth value True and Q has Truth value False. Other wise (PQ) is True

Conclution: From the above truth table if we observe the last TWO columns which are Not Equivalent.

So the given argument form is Invalid