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Proof step 14 from proof eq-pos (previous step)

i0 ^ i2 * i0 = i0 ^ i * i0, ~2 < i, ~2 < i2, 1 < i0, 2 ≤ i0, 0 ≤ i2, ~1 ≤ i2, 0 ≤ i, ~1 ≤ i, ∀ i1, i0, j. i→n(i1) < i→n(i2 +1) → i0 ^ i1 = i0 ^ j ∧ 1 < i0 ∧ 2 ≤ i0 ∧ 0 ≤ i1 ∧ 0 ≤ j → i1 = j ⊦ i2 = i

User applied rule apply induction

And obtained the following new goals

Proof step 15:
i0 ^ i2 * i0 = i0 ^ i * i0, 1 < i0, ~2 < i2, ~2 < i, ~1 ≤ i, 0 ≤ i, ~1 ≤ i2, 0 ≤ i2, 2 ≤ i0, ∀ i1, i0, j. i→n(i1) < i→n(i2 +1) → i0 ^ i1 = i0 ^ j ∧ 1 < i0 ∧ 2 ≤ i0 ∧ 0 ≤ i1 ∧ 0 ≤ j → i1 = j ⊦ i2 = i, i0 ^ i2 = i0 ^ i ∧ 1 < i0 ∧ 2 ≤ i0 ∧ 0 ≤ i2 ∧ 0 ≤ i

Rule argument

subst: [i1, i0, j] → [i2, i0, i]

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s)% ⊦ 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, i0, i