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

⊦ 0 ≤ i ∧ 0 ≤ j → (((i + 1) * (i + 2)) / 2 < ((j + 1) * ((j + 1) + 1)) / 2 ↔ i < j)

User applied rule cut formula

And obtained the following new goals

Proof step 4:
(i + 1) * (i + 2) = (i + 1) * ((i + 1) + 1) ⊦ 0 ≤ i ∧ 0 ≤ j → (((i + 1) * (i + 2)) / 2 < ((j + 1) * ((j + 1) + 1)) / 2 ↔ i < j)

Rule argument

fma: (i + 1) * (i + 2) = (i + 1) * ((i + 1) + 1)

Additional information

Used simplifier rules: add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: (i + 1) * (i + 2) = (i + 1) * ((i + 1) + 1)