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# discrete maths

show that (p-> q) -> r  and p -> (q -> r) are not logically equivalent

Posted in Math, asked by [email protected], 6 years ago. 1777 hits.

## 0

we have to show, (p->q)->r !=p->(q->r)

to  show these are not equivalent, we have to find one assignment of the truth values for p,q and r

Assume that, p=F

q=T

and   r=F

now L.H.S ,p->q=F->T=F

(P->q)->r=F->F=F

R.H.S,  q->r =T->F=F

p->(q->r)=F->F=T

So.L.H.S != R.H.S => (p->q)->r != p->(q->r)

Venkat Nagendra Thati - 6 years ago
Ask Venkat Nagendra Thati for further help.

## 0

P -> (q-> r)                                                               (p ->q) -> r

 P q r (p  ->q) (p ->q) ->r T T T T T T T F T F T F T F T T F F F T F T T T T F T F T F F F T T T F F F T F

 P q r (q ->r) P ->  (q ->r) T T T T T T T F F F T F T T T T F F T T F T T T T F T F F T F F T T T F F F T T

Ammara Hassan - 6 years ago
Ask Ammara Hassan for further help.
- Just now

## 0

Just now
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