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Proof step 14 from proof gcd-one-eq (previous step)

0 < gcd(i2, j2), j2 / gcd(i2, j2) = 1, j2 < 0, 0 ≤ gcd(i2, j2) ⊦ gcd(i2, j2) = j2

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-div inst: name: div-elim-negpos seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [j2, gcd(i2, j2), i, j] flags:

Additional information

Used simplifier rules: one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0); (elim) Used terms: j2, gcd(i2, j2), i, j