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Proof step 6 from proof prime-odd (previous step)

j % 2 = 0, 0 < j, prime(j) ⊦ j = 2

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: int-div name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [2, j] flags: <all paths>

Additional information

Used simplifier rules: prime-two: ⊦ prime(2 * j) ↔ j = 1;% ⊦ i * j = i ↔ i = 0 ∨ j = 1; (ws)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2, j