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Proof step 10 from proof gcd-divides-both-h (previous step)

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

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [j1, i1, j0] → [i, i0, j0]

Additional information

Used simplifier rules: gcd-com: ⊦ gcd(i, j) = gcd(j, i);%gcd-same-01: ⊦ 0 < j → gcd(j, j) = j;% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i ≤ i; (s) Used terms: i, i0, j0