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Proof step 2 from proof minus-odd (previous step)

odd(j), 0 < j ⊦ i ^ j + ~ i ^ j = 0

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:
∃ j0. j = j0 + j0 + 1, 0 < j ⊦ i ^ j + ~ i ^ j = 0

Rule argument

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

Additional information

Used terms: j