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Proof step 20 from proof exdivmod (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 21:
∃ 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

pos: left 1

Additional information

Used simplifier rules: neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1