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Proof step 5 from proof square-ls-04 (previous step)

m * m + m0 < n * n, n ≤ m, m0 ≤ 2 * m ⊦ n = 0

User applied rule cut formula

And obtained the following new goals

Proof step 6:
m * m + m0 < n * n, m0 ≤ 2 * m, n ≤ m, n * n ≤ m * m ⊦ n = 0

Rule argument

fma: n * n ≤ m * m

Additional information

Used simplifier rules: square-le-02: ⊦ m * m ≤ n * n ↔ m ≤ n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m ≤ n → (n < m ↔ false); (s) Used formulas: n * n ≤ m * m