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Proof step 34 from proof even-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: pot2-odd-02: ⊦ n / 2 * 2 + 1 = n ↔ odd(n);%oe: ⊦ ¬ odd(n) ↔ even(n);%Even: ⊦ even(n) ↔ (∃ m. m + m = n);%oe-02: ⊦ odd(n + 1) ↔ even(n);%meven-02: ⊦ even(2 * m);%two-succ-05: ⊦ 2 * m ≠ 2 * n + 1; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%add-num-01: n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); from specification nat%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%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