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Proof step 1 from proof square-ls-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs