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Best answer
4 votes
4 votes

[p → (q ∨ r)] ∧ (~q)} → (p → r)
= ((p' + q + r)q')' + p' + r
= (p'q' + rq')' + p' + r
= (p + q)(r' + q) + p' + r
= pr' + pq + qr' + q + p' + r
=pr' + (pq + qr' +q) + p' + r = (pr' + p') + q(p + r' +1) + r
= p' +r' + q + r
=p' + q + (r + r')
= p' +
 q + T
= TAUTOLOGY

 

selected by
5 votes
5 votes
implication is false when lhs is true and rhs is false

so rhs p->r is false if p is true and r is false (we assume RHS is false)

coming to lhs it is conjunction of two terms

-q is true  that means q is false

p->(q or r)    p is true q and r are false  so T->F is false

so lhs is false and false that gives false

F->F is true

so it is tautology(bcoz if rhs is false lhs is also false )

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