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Proof step 1 from proof uint2bits-2potrec (conclusion)

0 ≤ i, 2 ^ i ≤ j ⊦ uint2bits(j) = uint2bits(j / 2 ^ i) + fillsgn(i→n(i + 1), uint2bits(j % 2 ^ i)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i, 2 ^ i ≤ j ⊦ uint2bits(j) = uint2bits(j / 2 ^ i) + fillsgn(i→n(i) + 1, uint2bits(j % 2 ^ i)).rest

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;