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Proof step 14 from proof odd-add (previous step)

n = m1 + m1 + 1 ∧ odd(n) ⊦ ∃ m0. 2 * m0 = m, 2 * m1 = n, ∃ m0. 2 * m0 = m + n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ ∃ m0. 2 * m0 = m, ∃ m0. 2 * m0 = 2 * m1 + m + 1

Rule argument

Additional information

Used simplifier rules: oe-02: ⊦ odd(n + 1) ↔ even(n);%meven-02: ⊦ even(2 * m);%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