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Proof step 10 from proof modmod-add-law-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(i + i0 % j) % j = (i + i0) % j + j, (i0 % j + i % j) % j = 2 * j + (i + i0) % j, 0 < j ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ 2 * i = 0 ↔ i = 0; (s)% ⊦ i * j = i ↔ i = 0 ∨ j = 1; (ws)% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; (ws)% ⊦ i = i + j ↔ j = 0; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ (* i0) * i + i = (i0 + 1) * i; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + i = 2 * i; (s)