the answer given is Tautology, but m' unable to proove?

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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 )

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 )

2

Good method. You have gone inside the formula- so all GATE questions on this topic will be very easy :)

0

by considering this, how can we say its true for all combinations of p q r boolean values, thus forming tautology, thus a single case enough?