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Proof step 1 from proof otwo-03 (conclusion)

⊦ ¬ odd(2)

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ odd(n) ↔ (∃ m. n = m + m + 1)

Rule argument

spec: nat-even inst: name: oex seq: ⊦ odd(n) ↔ (∃ m. n = m + m + 1) subst: [n] → [2] flags: current, []

Additional information

Used simplifier rules: Odd: ⊦ odd(n) ↔ ¬ even(n);%oe-01: ⊦ odd(n) → ¬ even(n);%two-02: ⊦ m + m = 2 * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-num-04: n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; from specification nat%two-one-01: ⊦ 2 * m ≠ 1; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: 2