Without constructing the Truth Table, prove that (p→q) →q=pvq
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Answer:
L.H.S. =p↔q
=(p→q)∧(q→p)
=(p∨∼q)∧(q∨∼p)
=((p∨∼q)∧q)∨((p∨∼q)∧∼p)
=((p∧q)∨(∼q∧q))∨((p∧∼p)∨(∼q∧∼p))
=(p∧q)∨(∼p∧∼q)
= R.H.S.
Hence proved.
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