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Proof step 3 from proof div-zero-03 (previous step)

i < j, 0 < j, 0 ≤ i ⊦ i / j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)

Rule argument

spec: <toplevel> 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: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ 0 ≤ j1 ∧ j1 < j → (i0 * j + j1 < j ↔ i0 ≤ 0); (s)% ⊦ i < j → ¬ j < i; (s)%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% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) Used terms: i, j, i0, j0