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Proof step 3 from proof pls (previous step)

n < n0, m0 + n0 ≤ m + n, m ≤ m0 ⊦

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

spec: nat-basic name: pls seq: m < m0, n < n0 ⊦ m + n + 1 < m0 + n0 subst: [m, m0, n, n0] → [n, n0, m, m0]

Additional information

Used simplifier rules: lels: ⊦ m < n ↔ ¬ n ≤ m;%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n);%ler: ⊦ m ≤ m;%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false);%fle: ⊦ m < n ∧ n ≤ n0 → m < n0;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic Used terms: n, n0, m, m0