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Proof step 2 from proof le-sum (previous step)

∀ n, n0, n1. n ≤ n0 ∧ n1 ≤ m0 → n + n1 ≤ n0 + m0 ⊦ n ≤ n0 ∧ n1 ≤ m0 +1 → n + n1 ≤ n0 + m0 +1

User applied rule simplifier

And obtained the following new goals

Proof step 3:
n ≤ n0, n1 ≤ m0 + 1, n0 + m0 + 1 < n + n1, ∀ n, n0, n1. n ≤ n0 ∧ n1 ≤ m0 → n + n1 ≤ n0 + m0 ⊦ n + n1 = 0

Rule argument

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2