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Proof step 2 from proof ls-add-not (previous step)

m0 + n0 + n1 < m + n, m ≤ m0, n ≤ n0 ⊦ m + n = 0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
m + n ≤ m0 + n, m0 + n0 + n1 < m + n, m ≤ m0, n ≤ n0 ⊦ m + n = 0

Rule argument

fma: m + n ≤ m0 + n

Additional information

Used simplifier rules: lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%lels-01: ⊦ ¬ m ≤ n ↔ n < m;%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used formulas: m + n ≤ m0 + n