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Proof step 11 from proof mod-pot-shift-pos (previous step)

i0 = i2 + 1, 0 < j, 0 < i0, ∀ i1, j, i. i→n(i1) < i→n(i0) → 0 < j ∧ 0 < i1 → (i % j) ^ i1 % j = i ^ i1 % j ∨ i1 = 0 ⊦ ((i % j) ^ i2 * i % j) % j = i ^ i0 % j, i0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
0 < j, 0 ≤ i2, ∀ i1, j, i. i→n(i1) < i→n(i2 + 1) → 0 < j ∧ 0 < i1 → (i % j) ^ i1 % j = i ^ i1 % j ∨ i1 = 0 ⊦ ((i % j) ^ i2 * i % j) % j = (i ^ i2 * i) % j, ~1 = i2

Rule argument

Additional information

Used simplifier rules: succ: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)