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Proof step 2 from proof uint2bits-len (previous step)

0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 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) = i→n(i + 2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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) = i→n(i + 2)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i ≤ i;% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ 1 < i ∧ 0 ≤ j → (i ^ j < i ^ j0 ↔ j < j0);% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)