Exercise 1.6 | Q 6.3 | Page 16
Using the truth table, verify
~(p → ~q) ≡ p ∧ ~ (~ q) ≡ p ∧ q
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
p | q | ~q | p→~q | ~(p→~q) | ~(~q) | p∧~(~q) | p∧q |
T | T | F | F | T | T | T | T |
T | F | T | T | F | F | F | F |
F | T | F | T | F | T | F | F |
F | F | T | T | F | F | F | F |
In the above table, entries in columns 5, 7, and 8 are identical.
∴ ~(p → ~q) ≡ p ∧ ~ (~ q) ≡ p ∧ q
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