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Proof step 2 from proof uint2bits-div2pot (previous step)

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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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))

Proof step 2 from proof uint2bits-div2pot (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information