Discreet math
posted by James .
[(pvq)^(p=>)^(q=>)]=>r[(pvq)^(p=>)^(q=>)]=>r

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.