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Proof step 3 from proof div-ls-modzero (previous step)

i % j = 0, i0 * j < i, 0 < j, i / j ≤ i0, 0 ≤ i ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
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;%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); (s)% ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ i * 0 = 0; (s) Used terms: j, i