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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i0 ^ i = i0 ^ j, 1 < i0, 2 ≤ i0, 0 ≤ i, 0 ≤ j ⊦ i = j

Rule argument

Additional information

Used simplifier rules: ⊦ ~1 < i ↔ 0 ≤ i; (s)%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs