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Question
Using truth table show that ∼ (p → ∼ q) ≡ p ∧ q
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Solution
∼ (P → ∼ q) ≅ P ∧ q
=`(e^x/(2e^y))=1/2[e^(x-y)]`
`dy/dx=1/2[e^x+e^y]`
Truth table
| P (1) | q(2) | ∼q(3) | p→∼q (4) | ∼(p→∼q)(5) | p^q(6) |
| T | T | F | F | T | T |
| T | F | T | T | F | F |
| F | T | F | T | F | F |
| F | F | T | T | F | F |
From the truth table, we get 5th and 6th columns are
identical
∴∼(p→∼q)≅ p∧q
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