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Using truth table prove that p ↔ q = (p ∧ q) ∨ (~p ∧ ~q).

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#### Solution

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

p | q | p ↔ q | ~p | ~q | p ∧ q | ~p ∧ ~q | (p ∧ q) ∨(~p∧~q) |

T | T | T | F | F | T | F | T |

T | F | F | F | T | F | F | F |

F | T | F | T | F | F | F | F |

F | F | T | T | T | F | T | T |

The entries in columns 3 and 8 are identical.

p ↔ q = (p ∧ q) ∨ (~p ∧ ~q).

#### Notes

[1 mark each for column 3 and column 8]

Concept: Logical Connective, Simple and Compound Statements

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