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Proof step 9 from proof add-div-mult (previous step)

j0 < j, 0 ≤ i0, 0 < j, 0 ≤ j0, 0 ≤ i1, (j0 + i0 % j) / j = 1 ∨ (j0 + i0 % j) / j = 0 ⊦ (i1 + i0 / j) + (j0 + i0 % j) / j = ((i1 * j + j0) + i0) / j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ i + j + i0 = (i + j) + i0
Proof step 11:
j0 < j, 0 ≤ i0, 0 < j, 0 ≤ j0, 0 ≤ i1, (j0 + i0 % j) / j = 1 ∨ (j0 + i0 % j) / j = 0 ⊦ (i1 + i0 / j) + (j0 + i0 % j) / j = (i1 * j + j0 + i0) / j

Rule argument

spec: int-basic2 inst: name: add-a seq: ⊦ i + j + i0 = (i + j) + i0 subst: [i0, j, i] → [i0, j0, i1 * j] flags: rewrite

Additional information

Used terms: i0, j0, i1 * j