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Proof step 11 from proof gcd-divides-both-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
0 < i, 0 < i0, j0 | gcd(i, i0), i0 ≤ i, 0 ≤ i, abs(i0 + i) ≤ 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: ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); (s)