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Proof step 6 from proof uint2bits-len-01 (previous step)

0 < j ∧ j ≠ 1, 0 < 2 ^ i, 2 * 2 ^ i ≤ j, j < 3 * 2 ^ i, 0 ≤ 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) + (j % 2 = 1)) = i→n(i + 3)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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

Additional information

Used simplifier rules: ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ 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 ';