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Proof step 2 from proof sub-ls (previous step)

n + (m - n) < m ⊦ m = 0

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-01: ⊦ ¬ m ≤ n ↔ n < m;%lep: ⊦ m ≤ m + n;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%succ: ⊦ m +1 = m + 1; from specification nat-basic2%l-05: ⊦ m + m0 < n → ¬ n + n0 < m; from specification nat-basic2 Used terms: m, n, n0