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Proof step 1 from proof less (conclusion)

n ≤ m, n1 < n ⊦ (m - n) + n0 + n1 < m + n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n1 + 1 < m, n < (m - n) + n1, n1 < n, m ≤ (m - n) + n1, n1 < m, n < m ⊦ m = 0, n = 0, m = n, m + n1 = n, m - n = 1

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m;%fle: ⊦ m < n ∧ n ≤ n0 → m < n0;%zero-02: ⊦ m ≤ n → (n - m = 0 ↔ n = m);%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n);%lels-03: ⊦ m < n → m ≤ n;%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1;%zero-01: ⊦ n < m → m - n ≠ 0;%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%add-base: ⊦ n + 0 = n; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%l-01: ⊦ m < m + n ↔ n ≠ 0; 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: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2