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Question
Using the truth table prove the following logical equivalence.
p → (q → p) ≡ ∼ p → (p → q)
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Solution
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| p | q | q → p | p → (q → p) | ∼ p | p → q | ∼ p → (p → q) |
| T | T | T | T | F | T | T |
| T | F | T | T | F | F | T |
| F | T | F | T | T | T | T |
| F | F | T | T | T | T | T |
The entries in columns 4 and 7 are identical.
∴ p → (q → p) ≡ ∼ p → (p → q)
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