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Proof step 1 from proof exdivmodpos (conclusion)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
0 < j, 0 ≤ i ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0
Proof step 3:
∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0 ⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)

Rule argument

spec: int-mult inst: name: auxexdivmod-01 seq: 0 < j, 0 ≤ i ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0 subst: [j, i] → [j, i] flags: current, []

Additional information

Used terms: j, i