Verify following compound statements with reasons for each steps.
a ) [(pvq)∧(p∧¬q)]vq⇔pvq
b ) (p→q) ∧[(¬q∧(rv¬q)]⇔¬(qvp)
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a) [(p v q) ∧ (p ∧ ¬q)] v q ⇔ p v q
p - q
T - T
T - F
F - T
F - F
¬q
F
T
F
T
(p v q)
T
T
T
F
(p ∧ ¬q)
F
T
F
F
(p v q) ∧ (p ∧ ¬q)
F
T
F
F
[(p v q) ∧ (p ∧ ¬q)] v q
T
T
T
F
Thus, [(p v q) ∧ (p ∧ ¬q)] v q ⇔ p v q is true.
b) (p → q) ∧ [¬q ∧ (r v ¬q)] ⇔ ¬(q v p)
p - q - r
F - F - F
F - F - T
F - T - F
F - T - T
T - F - F
T - F - T
T - T - F
T - T - T
(p → q)
F
F
F
F
T
T
F
F
¬q
T
T
F
F
T
T
F
F
(r v ¬q)
T
T
F
T
T
T
F
T
¬q ∧ (r v ¬q)
T
T
F
F
T
T
F
F
(p → q) ∧ [¬q ∧ (r v ¬q)]
F
F
F
F
T
T
F
F
(q v p)
F
F
T
T
T
T
T
T
¬(q v p)
T
T
F
F
F
F
F
F
Thus, (p → q) ∧ [¬q ∧ (r v ¬q)] ⇔ ¬(q v p) is not possible.
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