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Proof step 12 from proof gcd-div-gcd-pos (previous step)

i = i1 * gcd(i, j), j = i2 * gcd(i, j) + j0, j0 < gcd(i, j), 0 < j, 0 < i, 0 < i0, 1 < i0, 0 < gcd(i, j), i0 | i1, i0 | i2, 0 ≤ i, 0 ≤ i1, 0 ≤ j, 0 ≤ i2, 0 ≤ j0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: gcd-div-rest seq: ⊦ abs(j0) < gcd(i, j) → (i = i1 * gcd(i, j) + j0 ↔ i = i1 * gcd(i, j) ∧ j0 = 0) subst: [j0, i1, j, i] → [j0, i2, i, j] flags: <all paths>

Additional information

Used simplifier rules: gcd-com: ⊦ gcd(i, j) = gcd(j, i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);% ⊦ 0 ≤ i → abs(i) = i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: j0, i2, i, j