Given the argument: A ⊃ ∼B / B ∨ A // ∼B ≡ A.
This argument is:
A) Invalid; fails in 4th line.
B) Invalid; fails in 2nd line.
C) Valid.
D) Invalid; fails in 1st line.
E) Invalid; fails in 3rd line.

Question

Quizplus · Accepted Answer

This is a valid argument that uses the principle of proof by contradiction.

1. A ⊃ ∼B  (premise)
2. B ∨ A  (premise) 
3. Assume ∼(∼B ≡ A) for the sake of contradiction. 
4. From 3, we derive ∼B ∧ ¬A by applying the definition of the biconditional and DeMorgan's Law. 
5. From 4, we derive ∼B by simplification. 
6. From 5 and 1, we derive ¬A by modus tollens. 
7. From 2 and 6, we derive B by disjunctive syllogism. 
8. From 5 and 7, we obtain a contradiction. 
9. Therefore, our assumption in 3 is false, and ∼B ≡ A (equivalence).

Given the Argument: a ⊃ ∼B / B ∨ a