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Proof step 5 from proof modmod-add-law-pos (previous step)

(i + i0 % j) % j = (i + i0) % j ∨ (i + i0 % j) % j = (i + i0) % j + j ∨ (i + i0 % j) % j = (i + i0) % j - j, (i0 % j + i % j) % j = (i0 % j + i) % j ∨ (i0 % j + i % j) % j = (i0 % j + i) % j + j ∨ (i0 % j + i % j) % j = (i0 % j + i) % j - j, 0 < j ⊦ (i % j + i0 % j) % j = (i + i0) % j ∨ (i % j + i0 % j) % j = (i + i0) % j + j ∨ (i % j + i0 % j) % j = (i + i0) % j - j

User applied rule case distinction

And obtained the following new goals

Proof step 6:
(i0 % j + i % j) % j = (i0 % j + i) % j + j, (i + i0 % j) % j = (i + i0) % j ∨ (i + i0 % j) % j = (i + i0) % j + j ∨ (i + i0 % j) % j = (i + i0) % j - j, 0 < j ⊦ (i % j + i0 % j) % j = (i + i0) % j ∨ (i % j + i0 % j) % j = (i + i0) % j + j ∨ (i % j + i0 % j) % j = (i + i0) % j - j
Proof step 14:
(i0 % j + i % j) % j = (i0 % j + i) % j - j, (i + i0 % j) % j = (i + i0) % j ∨ (i + i0 % j) % j = (i + i0) % j + j ∨ (i + i0 % j) % j = (i + i0) % j - j, 0 < j ⊦ (i % j + i0 % j) % j = (i + i0) % j ∨ (i % j + i0 % j) % j = (i + i0) % j + j ∨ (i % j + i0 % j) % j = (i + i0) % j - j

Rule argument

pos: left 2

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs