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Proof step 8 from proof two-loup-02 (previous step)

m + n = (2 * m1 + 2) + n0, m1 < n, n0 < 2, m + 1 < n, m1 < m, m < n, m1 + 1 < n ⊦ m + 1 = n

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 9:
¬ m0 < n0, n < m ⊦ n + n0 < m + m0

Rule argument

spec: nat-basic name: less seq: ¬ m0 < n0, n < m ⊦ n + n0 < m + m0 subst: [m0, n0, n, m] → [m, m1 + 1, m1 + 2, n]

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)%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)%n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0; (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m ≠ 1 → (m < 2 ↔ m = 0); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%n < n0 ⊦ m + (* n) < m0 + (* n0) ↔ m < m0 + (n0 - n); (s) Used terms: m, m1 + 1, m1 + 2, n