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Proof step 2 from proof auxexdivmod (previous step)

0 < j, i < (i0 + i1) * j, 0 ≤ i1, 0 ≤ i, 0 ≤ i0, i0 * j ≤ i ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
i < (i0 + i2 +1) * j, 0 < j, i0 * j ≤ i, 0 ≤ i0, 0 ≤ i, 0 ≤ i2 +1, ∀ j, i, i0. 0 < j ∧ i < (i0 + i2) * j ∧ 0 ≤ i2 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ i0 * j ≤ i → (∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0) ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0
Proof step 14:
i < (i0 + i2 -1) * j, 0 < j, i0 * j ≤ i, 0 ≤ i0, 0 ≤ i, 0 ≤ i2 -1, ∀ j, i, i0. 0 < j ∧ i < (i0 + i2) * j ∧ 0 ≤ i2 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ i0 * j ≤ i → (∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0) ⊦ ∃ 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;;% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ j → (j < i ↔ false); (as)