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March 24, 2017

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[(pvq)^(p=>)^(q=>)]=>r[(pvq)^(p=>)^(q=>)]=>r

  • Discreet math - ,

    If you need a proof of the proposition/identity, please first double check the expressions.

    [(pvq)^(p=>)^(q=>)]=>r[(pvq)^(p=>)^(q=>)]=>r

    Not sure if something is missing where expression is indicated in bold. Perhaps there are other problems.

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