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

0 ≤ i, 0 ≤ j ⊦ 2 ^ i = 2 ^ j ↔ i = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
2 ^ i = 2 ^ j, 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