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Proof step 2 from proof gcd-divides-both-h (previous step)

0 ≤ i ∧ i < j ∧ j0 | gcd(i, j), ∀ j1, i, j0. i→n(j1) < i→n(j) → 0 ≤ i ∧ i < j1 ∧ j0 | gcd(i, j1) → j0 | j1 ∧ j0 | i ⊦ j0 | j ∧ j0 | i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i < j, j0 | gcd(i, j), 0 ≤ i, 0 < j, ∀ j1, i, j0. i→n(j1) < i→n(j) → 0 ≤ i ∧ i < j1 ∧ j0 | gcd(i, j1) → j0 | j1 ∧ j0 | i ⊦ j0 | j ∧ j0 | i

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs