Use premise syllogism

(P→Q)∧(Q→R)？ P→R ①

(R→Q)∧(Q→P)？ R→P ②

Then use the eternal truth:

(P？ Q)∧(Q？ r)

(P→Q)∧(Q→P) ∧ (Q→R) ∧ (R→Q)

((P→Q)∧(Q→R)) ∧ ((R→Q)∧(Q→P))

(P→R) ∧ (R→P) Simplified formula and premise syllogism ① ②

p？ rare

therefore

(P？ Q)∧(Q？ r)？ p？ rare