DHARM
PROPOSITIONAL LOGIC 127
Sol.
Assume, S : Stephen loves Joyce
M : Matrye is happy
P : parents are happy
L : Shalezi will be happy
J : Stephen marries with Joyce
Then, symbolic representations of the statements are,
- S ∧ (~ M ∧ P)
- J → (L ∧ M)
- S → J
- S → (L ∧ M) 3 & 2, HS
- ~ S ∨ (L ∧ M) 4, Imp
- (~ S ∨ L) ∧ (~ S ∨ M) 5, Dist
- (~ S ∨ M) ∧ (~ S ∨ L) 6, Comm
- (~ S ∨ M) 7, Simp
- ~ (S ∧ ~ M) 8, DeM
- (S ∧ ~ M) ∧ P 1, Assoc
- (S ∧ ~ M) 10, Simp
- (S ∧ ~ M) ∧ ~ (S ∧ ~ M) 11 & 9, Add
Since we obtain a contradiction therefore given statements are inconsistent.
Example 5.25. Prove that the formula B ∨ (B → C) is a tautology.
Sol. Apply method of contradiction and assume that negation of formula is true. Thus,
We have
/∴ B ∨ (B → C)
- ~ (B ∨ (B → C)) IP (Indirect proof)
- ~ B ∧ ~ (B → C) 1, DeM
- ~ B 2, Simp
- ~ (B → C) ∧ ~ B 2, Comm
- ~ (B → C) 4, Simp
- ~ ( ~ B ∨ C) 5, Imp
- ~ ~ B ∧ ~ C 6, DeM
- ~ ~ B 7, Simp
- ~ ~ B ∧ ~ B 9 & 3, Conj/Add
Since, we get a contradiction hence deduction process stops. Hence the assumption ne-
gation of conclusion is false. Therefore, Formula is true or tautology.
Example 5.26. Prove that
/∴ ((A → B) ∧ (B → C)) → (A → C) is a tautology.
Sol. Since formula is of implication form, hence we use method of conditional proof. So
we shall include antecedent part as an additional premise and (A → C) is the only conclusion.
Still we have the conclusion is of implicative type so apply again method of conditional proof.
