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Proof step 2 from proof exdivmodpos (previous step)

0 < j, 0 ≤ i ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j

User applied rule apply lemma

And obtained the following new goals

Proof step 3:
0 < j, 0 ≤ i ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0
Proof step 4:
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

Rule argument

spec: <toplevel> name: auxexdivmod-01 seq: 0 < j, 0 ≤ i ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0 subst: [j, i] → [j, i]

Additional information

Used terms: j, i