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Question : 123 of 150
Marks:
+1,
-0
Solution:
Let r1 be the statement ∼(p⇔∼q). Then,
r1=∼(p⇔∼q)
⇒r1=∼{(∼p∨∼q)∧(p∨q)}
[∵p⇔q=(p∨∼q)∧(q∨∼p)]
⇒r1=∼{∼(p∧q)∧(p∨q)}
⇒r1=(p∧q)∨(∼p∧∼q)
⇒r1={(p∧q)∨∼p}∧{(p∧q)∨∼q}
⇒r1={(p∨∼p)∧(q∨∼p}∧{(p∨∼q)
⇒r1={t∧(q∨∼p)}∧{(p∨∼q)∧t}
⇒r1=(q∨∼p)∧(p∨∼q)
⇒r1=(p⇔q)
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