Answer is (1) (p ∧ (p → q)) → q
| p | q | p→q | p ∧ (p → q) | (p ∧ (p → q)) → q | q → p ∧(p → q) | p ∧ q | p ∨ (p ∧ q) | p ∨ q | p ∧ (p ∨ q) |
| T | T | T | T | T | T | T | T | T | T |
| T | F | F | F | T | T | F | T | T | T |
| F | T | T | F | T | F | F | F | T | F |
| F | F | T | F | T | F | F | F | F | F |