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Proof step 21 from proof exdivmod (previous step)

0 < i, j < 0, i ≤ 0 → i + j1 = j * i1 ∧ 0 ≤ i1 ∧ 0 ≤ j1 ∧ j1 + j < 0, 0 ≤ i ∧ 0 < ~ j → (∃ i0, j0. i = i0 * ~ j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < ~ j) ⊦ 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 22:
j < 0, 0 < i, i ≤ 0 → i + j1 = j * i1 ∧ 0 ≤ i1 ∧ 0 ≤ j1 ∧ j1 + j < 0, ∃ i0, j0. i = i0 * ~ j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < ~ j ⊦ j = 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

pos: left 4

Additional information

Used simplifier rules: ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s)