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Proof step 4 from proof gcd-rec-sub (previous step)

j = i0 + i, 0 < j, i < j, 0 < i, 1 < j ⊦ gcd(i, j) = gcd(i, i0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
1 < i0 + i, 0 < i0, 0 < i ⊦ gcd(i, i0 + i) = gcd(i, i0)

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ 0 < i ∧ 0 < j → 0 < i + j; (s)%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)