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

0 < j ∧ j ≠ 1, 2 ^ i ≤ j, 0 ≤ i, ∀ 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 / 2) + (j % 2 = 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j, 2 ^ i ≤ j, 0 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 → uint2bits(j0 / 2 ^ i) = butlastn(i→n(i), uint2bits(j0)) ⊦ j = 1, uint2bits(j / 2 ^ i) = butlastn(i→n(i), uint2bits(j / 2) + (j % 2 = 1)')

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar = bits + boolvar ';