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Proof step 10 from proof div-minus-onediff-01 (previous step)

j0 < 0, 0 < j ⊦ ((j0 - 1) * j + 1) / j = j0, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ i % j = 0 ∧ i0 % j = 0 ∧ 0 ≠ j → (i + i0) / j = i / j + i0 / j
Proof step 12:
0 < j, j0 < 0 ⊦ ((j0 - 1) * j + 1) / j = j0, j = 0, ((j0 - 1) * j) % j = 0 ∧ 1 % j = 0 ∧ 0 ≠ j
Proof step 20:
((j0 - 1) * j) % j = 0, 1 % j = 0, 0 < j, j0 < 0 ⊦ ((j0 - 1) * j) / j + 1 / j = j0, 0 = j

Rule argument

spec: <toplevel> name: div-add-both seq: ⊦ i % j = 0 ∧ i0 % j = 0 ∧ 0 ≠ j → (i + i0) / j = i / j + i0 / j subst: [j, i0, i] → [j, 1, (j0 - 1) * j] flags: <all paths>

Additional information

Used terms: j, 1, (j0 - 1) * j