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Proof step 11 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) -1, uint2bits(j / 2))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
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

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ n -1 = n - 1;