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Proof step 12 from proof two-loup-04 (previous step)

m + n = (2 * m1 + 2) + n0, m < m1, n0 < 2, m + 1 < n, m < n, n < m1, 2 + m < m1, m + 1 < m1, 2 * m1 ≤ m + n ⊦ m + 1 = 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] → [2 * m1, m + n, m, n]

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 < m + n0 ↔ n < n0; (s)% ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; (ws)% ⊦ m < n → ¬ n < m; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: 2 * m1, m + n, m, n