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Proof step 5 from proof mod-neg (previous step)

i % j = 0, 0 < j, i < j, i < 0 ⊦ j = i |%| j + ~ i |%| j, i |%| j = 0

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] → [i, j, i0, j0] flags:

Additional information

Used simplifier rules: mod-mult-01: ⊦ 0 < j → (i * j) |%| j = 0;%mod-mult-03: ⊦ 0 < j → ~(i * j) |%| j = 0;%ax-04: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%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: i, j, i0, j0