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Proof step 58 from proof gcd-rec-sub-03 (previous step)

0 < i, i ≥ i0 + i ⊦ gcd(i, i0 + i) = gcd(i0 + i, i0), i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 59:
0 < i, i0 < i, i0 ≤ 0 ⊦ gcd(i, i0 + i) = gcd(i0 + i, i0), i = 0

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i < i + j ↔ 0 < j; (s)% ⊦ i ≥ j ↔ ¬ i < j; (s)