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Proof step 2 from proof add-ls-right2-le (previous step)

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

User applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: m0 + n0 ≤ m0 + n1

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: m0 + n0 ≤ m0 + n1