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Propositional Logic set 2

Q6. [p ˄ (p → q) ˄ (q → r) → r] is equivalent to
a) T b) F c) R d) ~R
Solution: let F1 = p
F2 = p → q
F3 = q → r
F4 = r
F1 ˄ F2 ˄ F3 → F4
⇔~ (F1 ˄ F2 ˄ F3) ∨ F4 [XY = ~X V Y]
⇔~F1 ∨ ~F2 ∨ ~F3 ∨ F4
⇔ ~ p ∨ ~ (p → q) ∨ ~ (q → r) ∨ r
⇔~p ∨ ~ (~p ∨ q) ∨ ~ (~q ∨ r) ∨ r
⇔~p ∨ (p ˄ ~q) ∨ (q ˄ ~r) ∨ r
⇔ (~p ∨ p) ˄ (~p ∨ ~q) ∨ (q ˄ ~r) ∨ r
⇔ T ˄ (~p ∨ ~q) ∨ (q ˄ ~r) ∨ r
⇔ (~p ∨ ~q) ∨ (q ˄ ~r) ∨ r
⇔ (~p ∨ ~q) ∨ ((r ∨ ~r) ˄ (r ∨ q))
⇔ (~p ∨ ~q) ∨ (T ˄ (r ∨ q))
⇔~p ∨ ~q ∨ r ∨ q
⇔~p ∨ r ∨ ~q ∨ q
⇔~p ∨ r ∨ T
⇔T

Sir, in the above solution we are giving higher precedence to implication over conjunction. But actually conjunction is having higher precedence over implication. Please explaine above solution and correct me if I am wrong.

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