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Proof step 1 from proof add-ls-right2-le (conclusion)

m < m0 + n0, n0 ≤ n1 ⊦ m < m0 + n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m < m0 + n0, n0 ≤ n1, m0 + n1 ≤ m ⊦ m0 + n0 = 0

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1