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Proof step 18 from proof exdivmod (previous step)

j < 0, 0 < i, i ≤ 0 → i + j1 = j * i1 ∧ 0 ≤ i1 ∧ 0 ≤ j1 ∧ j1 + j < 0 ⊦ j = 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

User applied rule insert lemma

And obtained the following new goals

Proof step 19:
⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 20:
0 ≤ i ∧ 0 < ~ j → (∃ i0, j0. i = i0 * ~ j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < ~ j), j < 0, 0 < i, i ≤ 0 → i + j1 = j * i1 ∧ 0 ≤ i1 ∧ 0 ≤ j1 ∧ j1 + j < 0 ⊦ j = 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

spec: int-mult inst: name: exdivmodpos seq: ⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j) subst: [i, j] → [i, ~ j] flags: current, []

Additional information

Used terms: i, ~ j