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:
| P |
q |
S1 = p v q |
S2 = ~p |
S1 ∧ S2 |
S = q |
S1 ∧ 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.
