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Proof step 25 from proof gcd-mult-pos (previous step)

1 < abs(j % i), 1 < j, 0 < i, i < j, 0 < j, 0 < i1, i < abs(j % i), 0 ≤ j, j ≤ abs(j % i), ∀ j0, i0, i2. i→n(j0 * i0 + i2 * i0) < i→n(j * i1 + i * i1) → 0 < i2 ∧ 0 < i0 ∧ 0 ≤ j0 → gcd(i2 * i0, j0 * i0) = gcd(i2, j0) * i0 ⊦ j % i = 0, gcd(i * i1, j % i * i1) = gcd(i, j % i) * i1, i = 0, i1 = 0

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] → [j % i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ 0 < i ∧ 0 ≤ j → (j ≤ j % i ↔ j < i); (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ 0 < j ∧ 0 ≤ i → ¬ i % j < 0; (s) Used terms: j % i