## Discussion Forum

Que. | In propositional logic P ⇔ Q is equivalent to (Where ~ denotes NOT): |

a. | ~ (P ˅ Q) ˄ ~ (Q ˅ P) |

b. | (~ P ˅ Q) ˄ (~ Q ˅ P) |

c. | (P ˅ Q) ˄ (Q ˅ P) |

d. | ~ (P ˅ Q) → ~ (Q ˅ P) |

Answer:(~ P ˅ Q) ˄ (~ Q ˅ P) |

sandeep16064 :(May 26, 2020)
PQ'+P'Q
Complement of this became result |

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