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Proof step 6 from proof uint2bits-len (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 7:
1 < 2 ^ (i + 1), 0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1), 0 < j, 0 < 2 ^ (i + 1), 0 ≤ 2 ^ (i + 1), ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1) → # uint2bits(j0) = i→n(i + 2) ⊦ 0 ≤ i - 1 ∧ 2 ^ (i - 1) ≤ j / 2 ∧ j / 2 < 2 ^ ((i - 1) + 1), # uint2bits(j / 2) + 1 = i→n(i + 2), j = 1
Proof step 20:
# uint2bits(j / 2) = i→n((i - 1) + 2), 1 < 2 ^ (i + 1), 0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1), 0 < j, 0 < 2 ^ (i + 1), 0 ≤ 2 ^ (i + 1), ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1) → # uint2bits(j0) = i→n(i + 2) ⊦ # uint2bits(j / 2) + 1 = i→n(i + 2), 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