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Proof step 1 from proof auxexdivmod-01 (conclusion)

0 < j, 0 ≤ i ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
i0 * j ≤ i, i < (i0 + i1) * j, 0 ≤ i0, 0 < j, 0 ≤ i, 0 ≤ i1 ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0
Proof step 3:
0 < j, 0 ≤ i ⊦ 0 * j ≤ i ∧ i < (0 + i) * j ∧ 0 ≤ 0, ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0

Rule argument

spec: int-mult inst: name: auxexdivmod seq: i0 * j ≤ i, i < (i0 + i1) * j, 0 ≤ i0, 0 < j, 0 ≤ i, 0 ≤ i1 ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0 subst: [i0, j, i, i1] → [0, j, i, i] flags: current, []

Additional information

Used terms: 0, j, i, i