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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
Venkat Nagendra Thati - 6 years ago
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                      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

 

 

 

P

 

q

 

r

 

(q ->r)

 

P ->  (q ->r)

  

T

 

T

 

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T

 

T

 

F

 

F

 

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T

 

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T

 

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F

 

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F

 

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T

 

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Ammara Hassan
Ammara Hassan - 6 years ago
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