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Proof step 4 from proof square-inject (previous step)

i * (i + 1) + 2 * i0 = j * (j + 1) + 2 * j0, i0 ≤ i, j0 ≤ j, 0 ≤ i0, 0 ≤ j, 0 ≤ j0 ⊦ i = j ∧ i0 = j0, i * (i + 1) ≤ j * (j + 1) + 2 * j0 ∧ j * (j + 1) + 2 * j0 < (i + 1) * (i + 2) ∧ 0 ≤ i

User applied rule insert equation

And obtained the following new goals

Rule argument

pos: left 1 flags: rotate, discard

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