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Proof step 1 from proof square-ls (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs