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Proof step 3 from proof div-add-nosucc (previous step)

i + j = i0 + 1, i % j = 0, 0 < j, 1 < j, 0 ≤ i ⊦ i0 / j = i / j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 5:
j ≠ 0, i + j = i0 + 1, ∃ j0. i = j0 * j, 0 < j, 1 < j, 0 ≤ i ⊦ i0 / j = i / j

Rule argument

spec: inst: name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used simplifier rules: add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j, i