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Proof step 9 from proof le-sum (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m0 < n1, n0 + m0 < n + n1, m0 + n0 + 1 < n + n1, n1 ≤ m0 + 1, n ≤ n0, ∀ n, n0, n1. n ≤ n0 ∧ n1 ≤ m0 → n + n1 ≤ n0 + m0 ⊦

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0;%le-def: ⊦ m ≤ n ↔ ¬ n < m;%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0;%succ-02: ⊦ m < n + 1 ↔ m ≤ n;%ler: ⊦ m ≤ m;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)