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Proof step 8 from proof less-mult-div-01 (previous step)

i * i1 = i0 * j + j0, i1 < j, i ≤ i0, 0 < i, 0 < j1, j0 < j, 1 < j, 0 < j, 0 < i0, 0 ≤ j0, 0 < i1 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0)
Proof step 10:
i * i1 = i0 * j + j0, ∃ i0. j = i1 + i0 ∧ 0 < i0, i ≤ i0, 0 < i, 0 < j1, j0 < j, 1 < j, 0 < j, 0 < i0, 0 ≤ j0, 0 < i1 ⊦

Rule argument

spec: int-basic2 inst: name: ex-ls-add seq: ⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0) subst: [j, i] → [j, i1] flags: rewrite

Additional information

Used terms: j, i1