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Proof step 11 from proof gcd-rec-sub (previous step)

0 < i ∧ i ≤ i0 + i, 0 < i0, 1 < i0 + i, 0 < i, 0 ≤ i0 + i ⊦ gcd(i, (i0 + i - i) % i) = gcd(i, i0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-add: ⊦ i + j - i = j; from specification int-abs