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Proof step 4 from proof oex (previous step)

∃ n0. n = n0 + n0 ∨ n = (n0 + n0)+1 ⊦ ∃ m. n = 2 * m + 1, ∃ m. 2 * m = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n = 2 * n0 ∨ n = 2 * n0 + 1 ⊦ ∃ m. n = 2 * m + 1, ∃ m. 2 * m = n

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; 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