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Proof step 3 from proof nummult-modle-01 (previous step)

i % j = 0, 1 < i, j < i, 0 < j, 0 < i ⊦ i * i0 ≤ j * j0 ↔ i / j * i0 ≤ j0

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;% ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i * 0 = 0; (s)% ⊦ 0 < i → (i * j ≤ 0 ↔ j ≤ 0); (s) Used terms: j, i