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asked in Mathematical Logic by Boss (5.4k points)   | 61 views
in second option

,we can write (i) p -> Q ^ R (universal instatiation)

can we write (ii) as P ^ S?

because then from  (ii),we will get P (iii)& S (iv) through simplification.

from (iii) and (i), P -> Q ^ R

                            P

we get Q^ R(modes ponens) (v)

from (v),we can get Q (vi) & R (vii)(simplification)

with  (vii) and (iv) i.e R and S,we get R^ S(conjunction)

correct me if i am wrong
For 2nd premise, its given existential quantifier.  So, P^S is true only for some instances.

alright,so how will we know that for which intances we need to mark 0 in our k-map..??

so,surely this k-map is wrong cuz here,i have marked 0 for every P and evry S.caan you tell at which one to mark 0..?

P^S is SOP term, right? But there is existential quantifier.

So, when p=1 and S=1, the entries would be 1 only sometimes.

Now, Lets ignore P^S. So, remaining entries ( excluding the 2 premises)would be 1 or dont care, right?

 

Now, lets look at the conclusion.

$\exists$x { r(x) ^ s(x) }   which corresponds to third column from left.

So, you want 3rd column to be all 1's atleast once or for some instances.

 

Lets go back to $\exists$x { P(x) ^ S(x) }

So, now mark all 1's for this function because there exists atleast one instance where both P(x) and S(x) is true.

Now,  for this instance, just combine the 1's with dont care.

 

So, 2nd statement is also valid. I think I gave wrong answer initially :)

yes..i got that..thankyou so much..:-)

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