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Proof step 50 from proof gcd-rec-sub-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
0 ≤ i ⊦ gcd(i, i0 + i) = gcd(i0 + i, i0), 0 < i ∧ i < i0 + i

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)