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Proof step 12 from proof gcd-divides-both-h (previous step)

0 < i, 0 < i0, j0 | gcd(i, i0), i0 ≤ i, 0 ≤ i, abs(i0 + i) ≤ 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 insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [i0 + i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i + j ≤ i ↔ j ≤ 0; (s)% ⊦ i < j → (j ≤ i ↔ false); (as)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j; (s)% ⊦ i + i = 2 * i; (s)%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i + j < j ↔ i < 0; (s)% ⊦ i < j → ¬ j < i; (s) Used terms: i0 + i