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Proof step 4 from proof modmod-add-law-pos (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
0 < j ⊦ (i + i0 % j) % j = (i + i0) % j ∨ (i + i0 % j) % j = (i + i0) % j + j ∨ (i + i0 % j) % j = (i + i0) % j - j
Proof step 6:
0 < 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, (i + i0 % j) % j = (i + i0) % j ∨ (i + i0 % j) % j = (i + i0) % j + j ∨ (i + i0 % j) % j = (i + i0) % j - 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

spec: <toplevel> name: modmod-add-case seq: 0 < j ⊦ (i + i0 % j) % j = (i + i0) % j ∨ (i + i0 % j) % j = (i + i0) % j + j ∨ (i + i0 % j) % j = (i + i0) % j - j subst: [j, i, i0] → [j, i, i0]

Additional information

Used terms: j, i, i0