Let us show that each of these conditional statements is a tautology by using truth tables.
a) (p ∧ q) → p
p0011q0101p∧q0001(p∧q)→p1111
Since the last column contains only 1, we conclude that this formula is a tautology.
b) p → (p ∨ q)
p0011q0101p∨q0111p→(p∨q)1111
Since the last column contains only 1, we conclude that this formula is a tautology.
c) ¬p → (p → q)
p0011q0101¬p1100p→q1101¬p→(p→q)1111
Since the last column contains only 1, we conclude that this formula is a tautology.
d) (p ∧ q) → (p → q)
p0011q0101p∧q0001p→q1101(p∧q)→(p→q)1111
Since the last column contains only 1, we conclude that this formula is a tautology.
e) ¬(p → q) → p
p0011q0101p→q1101¬(p→q)0010¬(p→q)→p1111
Since the last column contains only 1, we conclude that this formula is a tautology.
f) ¬(p → q) → ¬q
p0011q0101p→q1101¬(p→q)0010¬q1010¬(p→q)→¬q1111
Since the last column contains only 1, we conclude that this formula is a tautology.
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