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Proof step 8 from proof uint2bits-div2pot (previous step)

¬ i→n(i) ≠ 0, 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)')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 < j, ∀ 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) = uint2bits(j / 2) + (j % 2 = 1)'

Rule argument

Additional information

Used simplifier rules: ⊦ 1 ≤ i ↔ 0 < i;% ⊦ butlastn(0, bits) = bits;% ⊦ i / 1 = i;% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i→n(i) = 0 ↔ i = 0;