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Proof step 1 from proof uint2bits-div2pot (conclusion)

0 ≤ i, 2 ^ i ≤ j ⊦ uint2bits(j / 2 ^ i) = butlastn(i→n(i), uint2bits(j))

User applied rule induction

And obtained the following new goals

Proof step 2:
0 ≤ i ∧ 2 ^ i ≤ j, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 → uint2bits(j0 / 2 ^ i) = butlastn(i→n(i), uint2bits(j0)) ⊦ uint2bits(j / 2 ^ i) = butlastn(i→n(i), uint2bits(j))

Rule argument

indvar: i→n(j) wfpred: <

Additional information

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