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Proof step 6 from proof square-interval (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i * (j + j0) = i * j + i * j0
Proof step 8:
2 * (j * (j + 1)) / 2 + 2 * j0 < 2 * ((i + 1) * (i + 2)) / 2, 0 ≤ j0, 0 ≤ j, 0 ≤ i, j0 ≤ j, (j * (j + 1)) / 2 + j0 < ((i + 1) * (i + 2)) / 2, (i * (i + 1)) / 2 ≤ (j * (j + 1)) / 2 + j0 ⊦ i = j, i * (i + 1) ≤ j * (j + 1) + 2 * j0 ∧ j * (j + 1) + 2 * j0 < (i + 1) * (i + 2)

Rule argument

spec: int-mult inst: name: dist-rsplit seq: ⊦ i * (j + j0) = i * j + i * j0 subst: [j0, j, i] → [j0, (j * (j + 1)) / 2, 2] flags: rewrite

Additional information

Used terms: j0, (j * (j + 1)) / 2, 2