[(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.
To evaluate the given expression [(pvq)^(p=>)^(q=>)]=>r[(pvq)^(p=>)^(q=>)]=>r, let's break it down step by step.
1. The given expression is [(pvq)^(p=>)^(q=>)]=>r[(pvq)^(p=>)^(q=>)]=>r. It seems there might be some missing parts of the expression. Could you please provide the full expression?
Please provide the complete expression, and I will be happy to help you evaluate it.