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Proof step 8 from proof gcd-divides-both-h (previous step)

0 < i0 + i, 0 < i0, j0 | gcd(i, i0), 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

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [j1, i1, j0] → [i0, i, j0]

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%div-zero-01: ⊦ i | 0;% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); (s)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i + j ≤ i ↔ j ≤ 0; (s)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + 0 = i; (s) Used terms: i0, i, j0