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Proof step 4 from proof eq-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information