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Proof step 8 from proof le-sum (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m0 < n1, n0 + m0 < n + n1, n ≤ n0, n1 ≤ m0 + 1, m0 + n0 + 1 < n + n1, ∀ n, n0, n1. n ≤ n0 ∧ n1 ≤ m0 → n + n1 ≤ n0 + m0 ⊦ n1 = 0, n + n1 = 0

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0;%lels-01: ⊦ ¬ m ≤ n ↔ n < m;%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0;%ler: ⊦ m ≤ m;%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2