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Proof step 7 from proof two-loup-07 (previous step)

m + n = m0 * 2 + n0, n0 < 2, n + 1 < m0, m < n, 2 * m0 ≤ m + n ⊦ m0 = 0, n = 0

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: two-02: ⊦ m + m = 2 * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%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%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n; from specification nat-basic2 Used terms: m, m0, n, m0