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B
a fallacy
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C
fallacy if q is false
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D
tautology if p is true
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Solution
The correct option is A a tautology [p∧(p→q)]→p=[p∧(¬p∨q)]→p=[(p∧¬p)∨(p∧q)]→p=(p∧q)→p=¬(p∧q)∨p=(¬p∨¬q)∨p=(¬p∨p)∨¬q=TRUE∨¬q=TRUE Hence, given statement is tautology