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Proof step 5 from proof prime-odd-01 (previous step)

(i0 + 1) % 2 = 0, prime(i0 + 1) ⊦ i0 % 2 = 0, 1 = i0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [2, i0 + 1] flags: rewrite

Additional information

Used simplifier rules: prime-two: ⊦ prime(2 * j) ↔ j = 1;%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: 2, i0 + 1