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Proof step 8 from proof two-loup-07 (previous step)

m + n = m0 * 2 + n0, m < n, n + 1 < m0, n0 < 2, 2 * m0 ≤ m + n ⊦

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: ⊦ m + m = 2 * m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m + m0 < n → ¬ n + n0 < m; (s)% ⊦ ¬ m < n → ¬ m + m0 < n; (s) Used terms: m, m0, n, m0