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Proof step 4 from proof odd-rec (
previous step
)
0 ≤ i0 ⊦ odd(i0 + 2) ↔ odd(i0)
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 5
:
⊦ odd(i) ↔ (∃ j. i = j + j + 1)
Proof step 6
:
0 ≤ i0 ⊦ (∃ j. i0 + 2 = j + j + 1) ↔ odd(i0)
Rule argument
spec: int-min inst: name: odd-def seq: ⊦ odd(i) ↔ (∃ j. i = j + j + 1) subst: [i] → [i0 + 2] flags: rewrite, current
Additional information
Used terms: i0 + 2