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Proof step 1 from proof oex (conclusion)

⊦ odd(n) ↔ (∃ m. n = m + m + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ∃ m. n = 2 * m + 1, ∃ m. 2 * m = n

Rule argument

Additional information

Used simplifier rules: Odd: ⊦ odd(n) ↔ ¬ even(n);%Even: ⊦ even(n) ↔ (∃ m. m + m = n);%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%two-succ-05: ⊦ 2 * m ≠ 2 * n + 1; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%two-02: ⊦ m + m = 2 * m; from specification nat-mult