# Discreet math

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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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