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Proof step 4 from proof exdivmodpos (previous step)

0 < j, 0 ≤ i, ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0 ⊦ ∃ 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 5:
0 < j, i < (i0 + 1) * j, 0 ≤ i, i0 * j ≤ i, 0 ≤ i0 ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j

Rule argument

pos: left 3

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