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Proof step 9 from proof int2bits-mod2-len (previous step)

i = i0 * 2 + j ∧ 0 ≤ j ∧ j < 2 ∧ 0 ≤ i0 ∧ 0 < 2 ∧ 0 ≤ i, 0 ≤ j ⊦ i ≤ 0 → # uint2bits(j) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 = i0 * 2 + j, j < 2, 0 ≤ i0, 0 ≤ j ⊦ # uint2bits(j) = 1

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);% ⊦ i ≤ j → (j ≤ i ↔ i = j);