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Proof step 6 from proof div-add-both (previous step)

i % j = 0 ⊦ (i + j0 * j) / j = i / j + j0, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)

Rule argument

spec: <toplevel> name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [j, i] flags: <all paths>

Additional information

Used simplifier rules: div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;% ⊦ i * j + i * j0 = i * (j + j0); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s) Used terms: j, i