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Proof step 55 from proof gcd-rec-sub-03 (previous step)

0 ≤ i ⊦ 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 56:
0 < i ⊦ gcd(i, i0 + i) = gcd(i0 + i, i0), i = 0, i < i0 + i

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)