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Proof step 6 from proof ls-add-not (previous step)

m0 + n0 + n1 < m + n, m0 + n ≤ m0 + n0, n ≤ n0, m ≤ m0, m + n ≤ m0 + n, m + n ≤ m0 + n0 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 7:
m0 + n0 + n1 < m + n, m + n ≤ m0 + n0, m + n ≤ m0 + n, m ≤ m0, n ≤ n0, m0 + n ≤ m0 + n0 ⊦ m0 + n0 + n1 < m0 + n0

Rule argument

fma: m0 + n0 + n1 < m0 + n0

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: m0 + n0 + n1 < m0 + n0