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Proof step 2 from proof exdivmod (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 4:
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) < abs(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