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Proof step 6 from proof two-loup-05 (previous step)

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

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] → [m0 + 1, n, m0 + 1, m]

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + m = 2 * m; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic Used terms: m0 + 1, n, m0 + 1, m