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Proof step 2 from proof less (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m, n, n0] flags: rewrite, current

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n;%lels-03: ⊦ m < n → m ≤ n;%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false);%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m, n, n0