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Proof step 3 from proof square-ls-02 (previous step)

m * (m + 1) < (n + 1) * ((n + 1) + 1) ↔ m < n + 1 ⊦ m * (m + 1) < (n + 1) * (n + 2) ↔ m < n + 1

User applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%succ: ⊦ (* m0) * m + m = (m0 + 1) * m;% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic