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Proof step 3 from proof add-div-mult (previous step)

0 < j, 0 ≤ i0, 0 ≤ i ⊦ (i / j + i0 / j) + (i % j + i0 % j) / j = (i + i0) / j

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
0 ≤ i, 0 ≤ i0, 0 < j ⊦ (i % j + i0 % j) / j = 1 ∨ (i % j + i0 % j) / j = 0
Proof step 5:
0 < j, 0 ≤ i, 0 ≤ i0, (i % j + i0 % j) / j = 1 ∨ (i % j + i0 % j) / j = 0 ⊦ (i / j + i0 / j) + (i % j + i0 % j) / j = (i + i0) / j

Rule argument

spec: <toplevel> name: mod-zero-one seq: 0 ≤ i, 0 ≤ i0, 0 < j ⊦ (i % j + i0 % j) / j = 1 ∨ (i % j + i0 % j) / j = 0 subst: [i, i0, j] → [i, i0, j]

Additional information

Used terms: i, i0, j