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Proof step 7 from proof uint2bits-len-01 (previous step)

0 ≤ i, j < 3 * 2 ^ i, 2 * 2 ^ i ≤ j, 0 < j, 0 < 2 ^ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ j0 < 3 * 2 ^ i ∧ 2 * 2 ^ i ≤ j0 ∧ 0 < 2 ^ i → # uint2bits(j0) = i→n(i + 3) ⊦ # uint2bits(j / 2) + 1 = i→n(i + 3), j = 1

User applied rule apply induction

And obtained the following new goals

Proof step 8:
0 ≤ i, j < 3 * 2 ^ i, 2 * 2 ^ i ≤ j, 0 < j, 0 < 2 ^ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ j0 < 3 * 2 ^ i ∧ 2 * 2 ^ i ≤ j0 ∧ 0 < 2 ^ i → # uint2bits(j0) = i→n(i + 3) ⊦ 0 ≤ i - 1 ∧ j / 2 < 3 * 2 ^ (i - 1) ∧ 2 * 2 ^ (i - 1) ≤ j / 2 ∧ 0 < 2 ^ (i - 1), # uint2bits(j / 2) + 1 = i→n(i + 3), j = 1
Proof step 23:
# uint2bits(j / 2) = i→n((i - 1) + 3), 0 ≤ i, j < 3 * 2 ^ i, 2 * 2 ^ i ≤ j, 0 < j, 0 < 2 ^ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ j0 < 3 * 2 ^ i ∧ 2 * 2 ^ i ≤ j0 ∧ 0 < 2 ^ i → # uint2bits(j0) = i→n(i + 3) ⊦ # uint2bits(j / 2) + 1 = i→n(i + 3), j = 1

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