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Proof step 1 from proof odd-def-mod (conclusion)

⊦ odd(i) ↔ i % 2 ≠ 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ¬ odd(i) ↔ (∃ j. i = j + j)
Proof step 3:
⊦ ¬(∃ j. i = j + j) ↔ i % 2 ≠ 0

Rule argument

spec: inst: name: not-odd-ex seq: ⊦ ¬ odd(i) ↔ (∃ j. i = j + j) subst: [i] → [i] flags: rewrite, current

Additional information

Used terms: i