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prove that (p^^q) ^^~(pvvq) is a contrad...

prove that `(p^^q) ^^~(pvvq)` is a contradiction.

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The correct Answer is:
`(p^^q)^^~(pvvq) -=(p^^q)^^(~p^^~q)`
`-=(p^^~p) ^^(q^^~q)`
`-=f^^f`
`-=f`
thus, `(p^^q)^^~(pvvq)` is fallacy, i.e., contradiction.
`(p^^q)^^~(pvvq) -=(p^^q)^^(~p^^~q)`
`-=(p^^~p) ^^(q^^~q)`
`-=f^^f`
`-=f`
thus, `(p^^q)^^~(pvvq)` is fallacy, i.e., contradiction.
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