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Which of the following statements is TRUE about the propositional logic formula

$S:\{(p→q)∧(¬q∨r)∧(r→s)\}→¬(p→s)$

$(A)$ S is a contradiction
$(B)$ S is satisfiable but not valid
$(C)$ S is valid
$(D)$ None of the above

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Option (B):

Take P = True  and Q = False

$(p\rightarrow q) = (T\rightarrow F) = F$

This will make whole L.H.S equal to FALSE, And ‘$F\rightarrow$ Anything’ will be TRUE

Thus, we can eliminate Option (A)

 

Now we’ll try to make this statement False

For that we need  L.H.S = T   &  R.H.S = F

Take P = Q = R = S = FALSE

$\left \{ (F\rightarrow F) \wedge (T\vee F)\wedge (F\rightarrow F) \right\}\rightarrow \sim(F\rightarrow F)$

Which will result in,   $T\rightarrow F$  =  $F$

Thus, we can also eliminate Options (C) 

And since we can get both True and False values for the given predicate, It is Satisfiable.

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