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Proof step 4 from proof q (previous step)

n0 = 1 ∨ m +1 +1 = n0 ⊦ n0 = 0, ∃ m0. m0 +1 = n0

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

fma: m +1 +1 = n0

Additional information

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