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

i < j, j0 | gcd(i, j), 0 ≤ i ⊦ j0 | j ∧ j0 | i

User applied rule induction

And obtained the following new goals

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

Rule argument

indvar: i→n(j) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j)