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Proof step 9 from proof gcd-divides-both-h (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 < i, 0 < i0, j0 | gcd(i, i0), i0 ≤ i, 0 ≤ i, ∀ j1, i1, j0. i→n(j1) < i→n(i0 + i) → i1 < j1 ∧ j0 | gcd(i1, j1) ∧ 0 ≤ i1 → j0 | j1 ∧ j0 | i1 ⊦ j0 | i0 ∧ j0 | i

Rule argument

Additional information

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