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Proof step 4 from proof div-eq-same-01 (previous step)

i / j = i ⊦ j = 0, i = 0, j = 1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
⊦ j < 0 ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ j < j0 ∧ 0 ≤ i0)
Proof step 6:
i / j = i ⊦ j < 0 ∧ i < 0, j = 0, i = 0, j = 1

Rule argument

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

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs Used terms: i, j, i0, j0