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Proof step 3 from proof exdivmodpos (previous step)

∃ 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)

User applied rule exists left

And obtained the following new goals

Proof step 4:
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

pos: left 1

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