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Proof step 4 from proof ex-two (previous step)

∃ m0. m0 + m0 = m ∨ m0 + m0 + 1 = m ⊦ ∃ n. m = 2 * n, ∃ n. m = 2 * n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 * m0 = m ∨ 2 * m0 + 1 = m ⊦ ∃ n. m = 2 * n, ∃ n. m = 2 * n + 1

Rule argument

Additional information

Used simplifier rules: two-02: ⊦ m + m = 2 * m;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2