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Proof step 9 from proof div-minus-onediff-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ i % j = 0 ∧ i0 % j = 0 ∧ 0 ≠ j → (i + i0) / j = i / j + i0 / j
Proof step 11:
¬(((j0 - 1) * j) % j = 0 ∧ 1 % j = 0 ∧ 0 ≠ j), 0 < j, j0 < 0 ⊦ j = 0, ((j0 - 1) * j + 1) / j = j0
Proof step 19:
((j0 - 1) * j) % j = 0 ∧ 1 % j = 0 ∧ 0 ≠ j, 0 < j, j0 < 0 ⊦ j = 0, ((j0 - 1) * j) / j + 1 / j = j0

Rule argument

spec: inst: 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: rewrite, current

Additional information

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