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Proof step 12 from proof uint2bits-div2pot (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 13:
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)) ⊦ 0 ≤ i - 1 ∧ 2 ^ (i - 1) ≤ j / 2, j = 1, uint2bits(j / 2 ^ i) = butlastn(i→n(i) - 1, uint2bits(j / 2)), i = 0
Proof step 24:
uint2bits((j / 2) / 2 ^ (i - 1)) = butlastn(i→n(i - 1), 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

Rule argument

subst: [j0, i] → [j / 2, i - 1]

Additional information

Used simplifier rules: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);%j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j);% ⊦ 0 < j → abs(j) = j;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < i → ¬ i ≤ i / 2; Used terms: j / 2, i - 1