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Proof step 23 from proof uint2bits-len-01 (previous step)

# 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
# uint2bits(j / 2) = i→n(2 + i), 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) ⊦ i→n(2 + i) + 1 = i→n(i + 3), j = 1

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + (j - i0) = i + j - i0;