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Proof step 2 from proof sqls (previous step)

m ≤ n ⊦ m + 1 < n * n ↔ n ≠ 0 ∧ n ≠ 1

User applied rule case distinction

And obtained the following new goals

Proof step 3:
n ≠ 0 ∧ n ≠ 1, m ≤ n ⊦ m + 1 < n * n

Rule argument

pos: right 1

Additional information

Used simplifier rules: sqzero: ⊦ m * m = 0 ↔ m = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2