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Proof step 4 from proof uint2bits-mod2pot-01 (previous step)

0 ≤ i ∧ 2 ^ i ≤ j * 2 ^ i + j0, 0 ≤ j0, 0 < j, 0 ≤ i, j0 < 2 ^ i ⊦ uint2bits((j * 2 ^ i + j0) / 2 ^ i) + fillsgn(i→n(i + 1), uint2bits((j * 2 ^ i + j0) % 2 ^ i)).rest = uint2bits(j) + fillsgn(i→n(i) + 1, uint2bits(j0)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 ^ i ≤ j * 2 ^ i + j0, 0 ≤ j0, 0 < j, 0 ≤ i, j0 < 2 ^ i ⊦ uint2bits((j * 2 ^ i + j0) / 2 ^ i) + fillsgn(i→n(i) + 1, uint2bits((j * 2 ^ i + j0) % 2 ^ i)).rest = uint2bits(j) + fillsgn(i→n(i) + 1, uint2bits(j0)).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;