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Proof step 24 from proof exdivmod (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 25:
⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 26:
0 ≤ ~ i ∧ 0 < j → (∃ i0, j0. ~ i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j), i < 0, 0 < j ⊦ 0 ≤ i, ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j

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