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Proof step 2 from proof square-ls-02 (previous step)

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

User applied rule insert equation

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i ∧ 0 ≤ j → (((i + 1) * (i + 2)) / 2 < ((j + 1) * ((j + 1) + 1)) / 2 ↔ i < j)

Rule argument

pos: left 1 flags: discard

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