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Proof step 3 from proof q (previous step)

n0 = 1 ∨ n0 = 0 ∨ (∃ m. m +1 +1 = n0) ⊦ n0 +1 = 1, n0 +1 = 0, ∃ m. m +1 +1 = n0 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 = 1 ∨ m +1 +1 = n0 ⊦ n0 = 0, ∃ m0. m0 +1 = n0

Rule argument

Additional information

Used simplifier rules: numeral-1: m ≠ 0 ⊦ n +1 = (* m) ↔ n = m -1;%disj-1: ⊦ 0 ≠ n +1;%inj-1: ⊦ n +1 = n0 +1 ↔ n = n0;