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Proof step 1 from proof sqr-03 (conclusion)

⊦ m + 1 < m * m ↔ 1 < m

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

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