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

0 ≤ i ⊦ bits2uint(uint2bits(i)) = i

User applied rule induction

And obtained the following new goals

Proof step 3:
0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → bits2uint(uint2bits(i0)) = i0 ⊦ bits2uint(uint2bits(i)) = i

Rule argument

indvar: i→n(i) wfpred: <

Additional information

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