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Proof step 5 from proof sqls (previous step)

m ≤ n, n * n ≤ m + 1 ⊦ n = 0, n = 1

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ n < n0 ∧ n0 < n1 → n +1 < n1

Rule argument

spec: nat-basic name: trans-01 seq: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1 subst: [n0, n, n1] → [n, m, n * n]

Additional information

Used simplifier rules: zero-01: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%sqr-06: ⊦ m * m ≤ m ↔ m ≤ 1;%sqr-07: ⊦ m * m ≤ m + 1 ↔ m ≤ 1;% ⊦ m +1 = m + 1; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m + n < m + n0 ↔ n < n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); (s)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) Used terms: n, m, n * n