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Proof step 11 from proof mod-ls-same-02 (previous step)

i < 0, j + i ≤ 0 ⊦ j = 0, i < i % j ↔ abs(j) = 1 ∨ ¬ 0 < abs(j) + i

User applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
⊦ j < 0 ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ j < j0 ∧ 0 ≤ i0)
Proof step 13:
i < 0, j + i ≤ 0 ⊦ j < 0 ∧ i < 0, j = 0, i < i % j ↔ abs(j) = 1 ∨ ¬ 0 < abs(j) + i
Proof step 16:
i = i0 * j + j0 ∧ j0 ≤ 0 ∧ j < j0 ∧ 0 ≤ i0 ∧ j < 0, i < 0, j + i ≤ 0 ⊦ j = 0, i < j0 ↔ abs(j) = 1 ∨ ¬ 0 < abs(j) + i

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 terms: i, j, i0, j0