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Proof step 1 from proof sqp (conclusion)

m0 ≠ 0, m0 ≠ 1, m0 * m0 ≤ m + 1 ⊦ m0 < m

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: mult-le-01: ⊦ m ≤ n → (n * n ≤ m + 1 ↔ n ≤ 1 ∧ m ≤ 1);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%one-02: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat