Use a truth table to show that (p->q)-> r and p->(q->r) are or are not logically equpvalent
Last edited by mr fantastic; Oct 17th 2009 at 02:08 PM. Reason: Changed post title
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Originally Posted by zap1231 Use a truth table to show that (p->q)-> r and p->(q->r) are or are not logically equpvalent No need for a whole talbe: when you give the value F to all p,q,r, you get in (p --> q) --> r the value F, whereas in p --> (q --> r) you get T. Tonio
Hello, zap1231! Use a truth table to show that and are or are not logically equivalent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .not equivalent
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