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Proof step 20 from proof uint2bits-len (previous step)

# 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# uint2bits(j / 2) = i→n(i) + 1, 0 ≤ 2 ^ (i + 1), 0 < 2 ^ (i + 1), 0 < j, j < 2 ^ (i + 1), 2 ^ i ≤ j, 0 ≤ i, 1 < 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) ⊦ j = 1, 2 + i→n(i) = i→n(i + 2)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + (j - i0) = i + j - i0;