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Proof step 2 from proof eq-pos (previous step)

i0 ^ i = i0 ^ j, 2 ≤ i0, 0 ≤ j, 0 ≤ i, 1 < i0 ⊦ i = j

User applied rule induction

And obtained the following new goals

Proof step 3:
i0 ^ i = i0 ^ j ∧ 2 ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ i ∧ 1 < i0, ∀ i1, i0, j. i→n(i1) < i→n(i) → i0 ^ i1 = i0 ^ j ∧ 2 ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ i1 ∧ 1 < i0 → i1 = j ⊦ i = j

Rule argument

indvar: i→n(i) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(i)