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Proof step 7 from proof gcd-divides-both-h (previous step)

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

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [j0, i1, j1] → [j0, i, i0]

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;%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lep-01: ⊦ i + j ≤ i ↔ j ≤ 0; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2 Used terms: j0, i, i0