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Proof step 5 from proof uint2bits-len (previous step)

0 < j ∧ j ≠ 1, j < 2 ^ (i + 1), 2 ^ i ≤ j, 0 ≤ i, ∀ 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) + (j % 2 = 1)) = i→n(i + 2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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

Additional information

Used simplifier rules: ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + boolvar = bits + boolvar ';