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Proof step 1 from proof eo-02 (conclusion)

⊦ even(n + 1) ↔ odd(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
2 * m = n + 1 ⊦ ∃ m0. n = 2 * m0 + 1

Rule argument

Additional information

Used simplifier rules: Even: ⊦ even(n) ↔ (∃ m. m + m = n);%oex: ⊦ odd(n) ↔ (∃ m. n = m + m + 1);%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%coef-17: ⊦ m * n + n = m1 * n ↔ ¬ ¬(m + 1 = m1 ∨ n = 0); 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