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Proof step 1 from proof square-ls-03 (conclusion)

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

User applied rule cut formula

And obtained the following new goals

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

Rule argument

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