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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information