Logic / Discrete Mathematics

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Show that "¬p Λ q is logically equivalent to p → ¬q" . I'm not sure if the conditional identity is correct, but what I did was I negated the whole second condition and it immediately became equal to the first? Please help and if possible, show steps or what laws I have to use. Thank you.

  • Logic / Discrete Mathematics -

    With simple expressions like these, the easiest proof is to make a truth table...unless it is specifically prohibited in the question.

    Also, there are precedence in the logical operators, just like + - * ÷.
    see
    http://en.wikipedia.org/wiki/Logical_connective

    The ¬ operator (which I write as ~ for simplicity) has the highest priority, in this order.

    Operator Precedence
    ¬ 1
    ∧ 2
    ∨ 3
    → 4
    ↔ 5

    There the first expression is interpreted as (~p)∧q which is not an identity to p->~q.

    However, writing the first expression as ~(p∧q) will give a truth table of
    TT F
    TF T
    FT T
    FF T
    identical to that of
    p->~q
    TT F
    TF T
    FT T
    FF T
    hence the identity
    ~(p∧q) ≡ p->~q

  • Logic / Discrete Mathematics -

    @MathMate Thank you so much! No wonder I was going nowhere even when I did the truth table, I had a mistake in translating the proposition. Thanks again.

  • Logic / Discrete Mathematics -

    You're welcome! :)

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