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Proof step 10 from proof le-sum (previous step)

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 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: anti seq: ⊦ m ≤ n → (n ≤ m ↔ m = n) subst: [m, n] → [m0 + 1, n1] flags: <all paths>

Additional information

Used simplifier rules: lels: ⊦ m < n ↔ ¬ n ≤ m;%succ-01: ⊦ m + 1 ≤ n ↔ m < n;%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false);%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%succ-02: ⊦ m < n + 1 ↔ m ≤ n;%ler: ⊦ m ≤ m;%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n);%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)%m < m0, n < n0 ⊦ m + n + 1 < m0 + n0; (s) Used terms: m0 + 1, n1