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Proof step 7 from proof odd-rec (previous step)

0 ≤ i0 ⊦ (∃ j. 2 * j = i0 + 1) ↔ odd(i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ odd(i) ↔ (∃ j. i = j + j + 1)
Proof step 9:
0 ≤ i0 ⊦ (∃ j. 2 * j = i0 + 1) ↔ (∃ j. i0 = j + j + 1)

Rule argument

spec: int-min inst: name: odd-def seq: ⊦ odd(i) ↔ (∃ j. i = j + j + 1) subst: [i] → [i0] flags: rewrite, current

Additional information

Used terms: i0