1.

Test the validity of the argument (S1 , S2 ; S), whereS1 ; p ∨ q, S2 : ~ p and S : q.

Answer»

In order to test the validity of the argument (S1 , S2 ; S), we first construct the truth table for the conditional statement.

S1 ∧ S2 → S i.e [(p ∨ q) ∧ ~p] →  q The truth table is as given below:

The truth table is as given below:

PqS1 = p v qS2 = ~pS1 ∧ S2S = qS1 ∧ S2 → S i.e. S1 ∧ S2 ​​​​​​​→ q
T
T
F
F		T
F
T
F		T
T
T
F		F
F
T
T		F
F
T
F		T
F
T
F		T
T
T
T

We observe that the last comumn of the truth table for S1 ∧ S2 ∧ S2 → contains T only.

Thus, S1 ∧ S2 →  S is a tautology.
Hence, the given argument is valid.


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