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Proof step 2 from proof square-nle (previous step)

0 ≤ j ⊦ (j * (j + 1)) / 2 ≤ j ↔ j < 2

User applied rule case distinction

And obtained the following new goals

Proof step 3:
(j * (j + 1)) / 2 ≤ j, 0 ≤ j ⊦ j < 2
Proof step 19:
j < 2, 0 ≤ j ⊦ (j * (j + 1)) / 2 ≤ j

Rule argument

pos: right 1

Additional information