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Proof step 1 from proof otwo-02 (conclusion)

⊦ odd(m + 2) ↔ odd(m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
2 * m0 = m + 1 ⊦ ∃ m1. m = 2 * m1 + 1

Rule argument

Additional information

Used simplifier rules: 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% ⊦ m * n + n = m1 * n ↔ m + 1 = m1 ∨ n = 0; (s)%n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + m = 2 * m; (s)