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Proof step 4 from proof q (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m0 = n + n ∨ m0 = (n + n) + 1 ⊦ ∃ n0. m0 + 1 = n0 + n0 ∨ m0 = n0 + n0

Rule argument

Additional information

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