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Proof step 6 from proof gcd-one-eq (previous step)

j2 / gcd(i2, j2) = 1, 0 ≤ gcd(i2, j2), j2 ≥ 0 ⊦ gcd(i2, j2) = j2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j2 / gcd(i2, j2) = 1, 0 ≤ gcd(i2, j2), 0 ≤ j2 ⊦ gcd(i2, j2) = j2

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≥ j ↔ ¬ i < j; (s)