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Proof step 2 from proof mod-succ-02 (previous step)

i % j + 1 = j, 0 ≤ i, 0 < j ⊦ (i + 1) % j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)

Rule argument

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

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs Used terms: i, j, i0, j0