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Proof step 6 from proof nummult-le-01 (previous step)

j < i, j < 0, 0 < i, j + 1 < i ⊦ i * i0 ≤ j ↔ i0 ≤ (j + 1 - i) / i, i = j + 1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j)
Proof step 8:
j + 1 - i = i1 * i + j0 ∧ j0 ≤ 0 ∧ abs(j0) < i ∧ j + 1 - i < 0, j < i, j < 0, 0 < i, j + 1 < i ⊦ i * i0 ≤ j ↔ i0 ≤ i1, i = j + 1

Rule argument

spec: int-div inst: name: div-elim2 seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j) subst: [i, j, i0, j0] → [j + 1 - i, i, i1, j0] flags: rewrite, current

Additional information

Used simplifier rules: sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-zero: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; from specification int-basic2 Used terms: j + 1 - i, i, i1, j0