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Proof step 11 from proof div-eq-zero-01 (previous step)

i / j = 0, i < 0, i < j ⊦ 0 = j, 0 < abs(j) + i

User applied rule insert elim lemma

And obtained the following new goals

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

Rule argument

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

Additional information

Used simplifier rules: ⊦ i * 0 = 0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i + 0 = i; (s)%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)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ 0 < j → abs(j) = j; (s) Used terms: i, j, i0, j0