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Proof step 20 from proof square-interval (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ i * (j + j0) = i * j + i * j0
Proof step 22:
0 < i * (i + 1), 0 < (i + 1) * (i + 2), j * (j + 1) + 2 * j0 < (i + 1) * (i + 2), (j * (j + 1)) / 2 + j0 < ((i + 1) * (i + 2)) / 2, 0 < j * (j + 1), 0 ≤ j, 0 ≤ j0, 0 ≤ i, j0 ≤ j, (i * (i + 1)) / 2 ≤ (j * (j + 1)) / 2 + j0, i * (i + 1) ≤ 2 * (j * (j + 1)) / 2 + 2 * j0 ⊦ (i * (i + 1)) % 2 = 1, ((i + 1) * (i + 2)) % 2 = 1, (j * (j + 1)) % 2 = 1, i = j, i * (i + 1) ≤ j * (j + 1) + 2 * j0

Rule argument

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

Additional information

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