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Proof step 9 from proof gcd-one-eq (previous step)

j2 ≠ 0, gcd(i2, j2) | j2 ∧ gcd(i2, j2) ≠ 0 → (j2 / gcd(i2, j2) = 1 ↔ j2 = gcd(i2, j2)), j2 / gcd(i2, j2) = 1, 0 ≤ gcd(i2, j2), 0 ≤ j2 ⊦ gcd(i2, j2) = j2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: gcd-zero-zero: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs