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Proof step 3 from proof less-mult-div-01 (previous step)

i1 < j, 0 < j1, 0 < j, 0 < i, 0 < (i * i1) / j, i ≤ (i * i1) / j ⊦

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 4:
⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 5:
0 < (i * i1) / j, 0 < i, 0 < j, 0 < j1, i1 < j, i ≤ (i * i1) / j, 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j) ⊦

Rule argument

spec: int-mult name: exdivmodpos seq: ⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j) subst: [i, j] → [i, j]

Additional information

Used terms: i, j