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Proof step 6 from proof square-ls-04 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 7:
m * m + m0 < n * n, n * n ≤ m * m, n ≤ m, m0 ≤ 2 * m ⊦ n = 0, m * m + m0 < m * m

Rule argument

fma: m * m + m0 < m * m

Additional information

Used simplifier rules: ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: m * m + m0 < m * m