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Proof step 4 from proof sqls (previous step)

m ≤ n, n * n ≤ m + 1 ⊦ n = 1, n = 0

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 5:
⊦ n < n0 ∧ n0 < n1 → n +1 < n1

Rule argument

spec: nat-basic1 inst: name: trans seq: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1 subst: [n0, n, n1] → [n, m, n * n] flags:

Additional information

Used simplifier rules: ls-07: ⊦ n0 < n → (n < m * n ↔ 1 < m);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%sqr-07: ⊦ m * m ≤ m + 1 ↔ m ≤ 1;%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%succ: ⊦ m +1 = m + 1; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%one-02: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: n, m, n * n