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Proof step 1 from proof sqr-01 (conclusion)

⊦ m < m * m ↔ 1 < m

User applied rule case distinction

And closed the goal!

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;%ls-04: ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m);%leone: ⊦ n ≠ 0 → (m * n ≤ n ↔ m ≤ 1);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%one-02: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1