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Proof step 6 from proof even-06 (previous step)

2 + m0 + m0 = 3 + n0 + n0, ∀ m. m + m ≠ n0 + n0 + 1, ∀ n. m0 + m0 ≠ 3 + n + n ∨ ¬(∀ m. m + m ≠ n + n + 1) ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 7:
3 = 2 + 1, 2 + m0 + m0 = 3 + n0 + n0, ∀ m. m + m ≠ n0 + n0 + 1, ∀ n. m0 + m0 ≠ 3 + n + n ∨ ¬(∀ m. m + m ≠ n + n + 1) ⊦

Rule argument

fma: 3 = 2 + 1

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m; Used formulas: 3 = 2 + 1