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Proof step 1 from proof auxexdivmod (conclusion)

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

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ i0, j, i. i0 * j ≤ i ∧ i < (i0 + i2) * j ∧ 0 ≤ i0 ∧ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i2 → (∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0), i0 * j ≤ i, i < (i0 + i2 +1) * j, 0 ≤ i0, 0 < j, 0 ≤ i, 0 ≤ i2 +1 ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0
Proof step 13:
∀ i0, j, i. i0 * j ≤ i ∧ i < (i0 + i2) * j ∧ 0 ≤ i0 ∧ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i2 → (∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0), i0 * j ≤ i, i < (i0 + i2 -1) * j, 0 ≤ i0, 0 < j, 0 ≤ i, 0 ≤ i2 -1 ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0

Rule argument

indvar: i1

Additional information

Used simplifier rules: int generated by 0, +1, -1;;%add-zero: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels-04: ⊦ i ≤ j → (¬ ¬ j < i ↔ false); from specification int-abs