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Proof step 1 from proof square-inject (conclusion)

⊦ 0 ≤ i ∧ 0 ≤ j ∧ 0 ≤ i0 ∧ 0 ≤ j0 → (((i + j) * (i + j + 1)) / 2 + j = ((i0 + j0) * (i0 + j0 + 1)) / 2 + j0 ↔ i = i0 ∧ j = j0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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