Thread: Truth table

Use a truth table to show that (p->q)-> r and p->(q->r) are or are not logically equpvalent

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