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

∀ n0. m0 * m0 + m0 < n0 * n0 + n0 ↔ m0 < n0 ⊦ 3 * m0 + m0 * m0 + 2 < n * n + n ↔ m0 + 1 < n

User applied rule cut formula

And obtained the following new goals

Proof step 4:
∀ n0. m0 * m0 + m0 < n0 * n0 + n0 ↔ m0 < n0 ⊦ n = 0, 3 * m0 + m0 * m0 + 2 < n * n + n ↔ m0 + 1 < n

Rule argument

fma: n = 0

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l-num-06: n0 < n ⊦ ¬ m + (* n) < (* n0); from specification nat Used formulas: n = 0