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Proof step 15 from proof mod-pot-shift-pos (previous step)

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 % j * i % j) % j = (i ^ i2 * i) % j, 0 = j, ~1 = i2

User applied rule apply induction

And obtained the following new goals

Proof step 16:
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 % j * i % j) % j = (i ^ i2 * i) % j, ~1 = i2, 0 = j, 0 < j ∧ 0 < i2
Proof step 19:
0 < j, 0 ≤ i2, (i % j) ^ i2 % j = i ^ i2 % j ∨ i2 = 0, ∀ 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 % j * i % j) % j = (i ^ i2 * i) % j, ~1 = i2, 0 = j

Rule argument

subst: [i1, j, i] → [i2, j, i]

Additional information

Used simplifier rules: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); (s)% ⊦ 0 ≤ i → abs(i) = i; (s)% ⊦ ¬ i < i; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: i2, j, i