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Proof step 5 from proof uint2bits-div2pot-01 (previous step)

0 ≤ i ∧ 2 ^ i ≤ j, 0 ≤ i, 0 < j, j % 2 ^ i = 0 ⊦ uint2bits(j) = butlastn(i→n(i), uint2bits(j)) + mklist(false, i→n(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j % 2 ^ i = 0, 0 < j, 0 ≤ i, 2 ^ i ≤ j ⊦ uint2bits(j) = butlastn(i→n(i), uint2bits(j)) + mklist(false, i→n(i))

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ i = i0 → i0 < j; (forward)