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Construct a truth table for (p ^ q) ↔ ~p

  • Math-geometry -

    the truth table for bi-conditional <-> is
    it is true when both p & q are true, or both are false.

    p q (p^q) <-> ~p
    T T T T T F F
    T F T F F F T
    F T F F T F F
    F F F F F F T

    Since in all cases, the proposition is false, it is a contradiction.

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