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Proof step 6 from proof even-mult (previous step)

(i * (2 * i0 + 1)) % 2 = 0 ⊦ i % 2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i % 2 = 0 ↔ ¬(∃ i0. i = 2 * i0 + 1)
Proof step 8:
(i * (2 * i0 + 1)) % 2 = 0 ⊦ ∀ i0. i ≠ 2 * i0 + 1

Rule argument

spec: <toplevel> name: even-not-ex seq: ⊦ i % 2 = 0 ↔ ¬(∃ i0. i = 2 * i0 + 1) subst: [i] → [i] flags: <all paths>

Additional information

Used terms: i