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    				| 1. | 
                                    prove that `(p^^q) ^^~(pvvq)` is a contradiction. | 
                            
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Answer» `(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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