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Proof step 1 from proof square-ls (conclusion)

⊦ m * (m + 1) < n * (n + 1) ↔ m < n

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ n. m0 * (m0 + 1) < n * (n + 1) ↔ m0 < n ⊦ m0 +1 * (m0 +1 + 1) < n * (n + 1) ↔ m0 +1 < n

Rule argument

indvar: m

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%rsucc-01: ⊦ (m + 1) * n = m * n + n;%nat freely generated by 0, +1;;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2