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

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

User applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: m0 + n1 ≤ m0 + n0

Additional information

Used simplifier rules: lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%ls-le-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 + n1 ≤ m0 + n0