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Proof step 1 from proof sqr-02 (conclusion)

⊦ m + 1 ≤ m * m ↔ 1 < m

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: sqr-01: ⊦ m < m * m ↔ 1 < m;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%ls-04: ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat