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Proof step 25 from proof uint2bits-div2pot (previous step)

i = i0 + 1, uint2bits((j / 2) / 2 ^ i0) = butlastn(i→n(i0), uint2bits(j / 2)), 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) - 1, uint2bits(j / 2)), i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
uint2bits((j / 2) / 2 ^ i0) = butlastn(i→n(i0), uint2bits(j / 2)), 0 < j, 2 ^ (i0 + 1) ≤ j, 0 ≤ i0, ∀ 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 ^ (i0 + 1)) = butlastn(i→n(i0), uint2bits(j / 2)), ~1 = i0

Rule argument

Additional information

Used simplifier rules: ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ 0 < i + 1 → i→n(i + 1) = i→n(i) + 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;