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Proof step 3 from proof square-ls-04 (previous step)

m * m + m0 < n * n, m0 ≤ 2 * m ⊦ m < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ m, m0 ≤ 2 * m, m * m + m0 < n * n ⊦ n = 0

Rule argument

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%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat