P: a is an odd number.

Q: A is divisible by 2.

R: a is an even number.

If a is an odd number, a cannot be divisible by 2: p→? q

If a is an even number, then a is divisible by 2: r → q.

If a is even, then a is not odd: r→? p

(p→? q)∧(r→q)

(? p∨？ q)∧(？ R∨q) becomes conjunctive disjunction.

(? p∨？ q∨(？ r∧r))∧((？ p∧p)∨q∨？ R) supplement

((? p∨？ q∨？ r)∧(？ p∨？ q∨r))∧((？ p∧p)∨q∨？ R) distribution method

(? p∨？ q∨？ r)∧(？ p∨？ q∨r)∧((？ p∧p)∨q∨？ R) association law

(? p∨？ q∨？ r)∧(？ p∨？ q∨r)∧((？ p∨q∨？ r)∧(p∨q∨？ R)) distribution law

(? p∨？ q∨？ r)∧(？ p∨？ q∨r)∧(？ p∨q∨？ r)∧(p∨q∨？ R) binding laws 1

Get Lord conjunctive normal form,

And r → p.

r∨？ P becomes conjunctive disjunction

r∨？ p ∨(q∧？ Q)

(? r∨？ p ∨q)∧(？ r∨？ p ∨？ Q) 2

Obviously, the conjunctive subformula in 2 is contained in 1.

Therefore, the reasoning is correct.

that is

(p→? q)∧(r→q)？ r→? p