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Proof step 3 from proof less-03 (previous step)

n + n0 < m ⊦ m = 0, m0 ≠ n1 ∧ ¬ n1 < m0, n + n0 + n1 ≠ m + m0 ∧ ¬ m + m0 < n + n0 + n1

User applied rule cut formula

And obtained the following new goals

Proof step 4:
n + n0 < m ⊦ m = 0, m0 = n1, n + n0 + n1 ≠ m + m0 ∧ ¬ m + m0 < n + n0 + n1, m0 ≠ n1 ∧ ¬ n1 < m0

Rule argument

fma: m0 = n1

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%l: ⊦ m + n < m + n0 ↔ n < n0;%ls-02: ⊦ m < n → ¬ n < m;%ref: ⊦ ¬ n < n; Used formulas: m0 = n1