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Proof step 2 from proof not-odd-ex (previous step)

⊦ ¬ odd(i) ↔ (∃ j. i = 2 * j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ odd(i) ↔ (∃ j. i = j + j + 1)
Proof step 4:
⊦ ¬(∃ j. i = j + j + 1) ↔ (∃ j. i = 2 * j)

Rule argument

spec: int-min inst: name: odd-def seq: ⊦ odd(i) ↔ (∃ j. i = j + j + 1) subst: [i] → [i] flags: rewrite, current

Additional information

Used terms: i