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Proof step 5 from proof gcd-divides-both-h (previous step)

i < j, j0 | gcd(i, j - i), 0 ≤ i, 0 < j, ∀ j1, i, j0. i→n(j1) < i→n(j) → 0 ≤ i ∧ i < j1 ∧ j0 | gcd(i, j1) → j0 | j1 ∧ j0 | i ⊦ j0 | j ∧ j0 | i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
j = i0 + i, i < j, j0 | gcd(i, i0), 0 ≤ i, 0 < j, ∀ j1, i, j0. i→n(j1) < i→n(j) → 0 ≤ i ∧ i < j1 ∧ j0 | gcd(i, j1) → j0 | j1 ∧ j0 | i ⊦ j0 | j ∧ j0 | i

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, i, i0] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, i, i0