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Math Help - Standard logical equivalences

  1. #1
    Junior Member
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    Standard logical equivalences

    Hi all. i've got a one question well i need to show that p -> (q ^ r) and (p -> q) ^ (p -> r) are logically equivalent.

    This is what i did ;

    let p be A and B be (q^r)
    A->B
    not(A or B) - implication laws
    not p or (q^r)
    (not p or q) ^ (not p or q) - De Morgan laws
    (not p or q) <=> p->q - implication laws
    (not p or r) <=> p->r - implication laws

    (p -> q) ^ (p -> r) <=> p -> (q ^ r)

    Please anyone let me know if this is all right.

    Regards
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Snowboarder View Post
    Hi all. i've got a one question well i need to show that p -> (q ^ r) and (p -> q) ^ (p -> r) are logically equivalent.

    This is what i did ;

    let p be A and B be (q^r)
    A->B
    not(A or B) - implication laws
    not p or (q^r)
    (not p or q) ^ (not p or q) - De Morgan laws
    (not p or q) <=> p->q - implication laws
    (not p or r) <=> p->r - implication laws

    (p -> q) ^ (p -> r) <=> p -> (q ^ r)

    Please anyone let me know if this is all right.

    Regards
    I think your basic idea is right, although the way you write it up seems unnecessarily messy to me. In particular I find it difficult to understand what introducing A and B gives you. Also, you probably meant to write that A\rightarrow B is equivalent to \neg A\vee B instead of \neg (A\vee B). And I don't see an application of De Morgan's laws either...

    Why not write the following:
    \begin{array}{lcll}<br />
p\rightarrow (q\wedge r) &\Leftrightarrow& \neg p \vee (q\wedge r) &\text{(implication law)}\\<br />
&\Leftrightarrow& (\neg p \vee q)\wedge (\neg p\vee r)& \text{(distributivity law)}\\<br />
&\Leftrightarrow& (p\rightarrow q)\wedge (p\rightarrow r)&\text{(implication law)}\end{array}
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, Snowboarder!

    Show that these are equivalent: . <br />
p \to (q \wedge r)\,\text{ and }\,(p \to q) \wedge (p \to  r)
    DeMorgan does not apply here . . .



    . . \begin{array}{ccc}p \to (q \wedge r) && \text{Given} \\ \\ \sim p \vee (q \wedge r) && \text{Implication Law} \\ \\ (\sim p \vee  q) \wedge (\sim p \vee q) && \text{Distributive Law} \\ \\ (p \to q) \wedge (p \to r) && \text{Implication Law }\end{array}

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