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Proof step 2 from proof uint2uint-inv (previous step)

0 < i ⊦ 2 ^ n→i(# uint2bits(i)) = bits2uint(bitsinvert(uint2bits(i))) + i + 1

User applied rule induction

And obtained the following new goals

Proof step 3:
0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → 2 ^ n→i(# uint2bits(i0)) = bits2uint(bitsinvert(uint2bits(i0))) + i0 + 1 ⊦ 2 ^ n→i(# uint2bits(i)) = bits2uint(bitsinvert(uint2bits(i))) + i + 1

Rule argument

indvar: i→n(i) wfpred: <

Additional information

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