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Proof step 6 from proof q (previous step)

m0 = n + n ∨ m0 = (n + n) + 1 ⊦ m0 = n + n, ∃ n0. m0 + 1 = n0 + n0 ∨ m0 = n0 + n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ ∃ n0. 2 + n + n = n0 + n0

Rule argument

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0;%eq: ⊦ m = m + n ↔ n = 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%even-06: ⊦ m + m ≠ n + n + 1;