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Proof step 20 from proof uint2bits-len-01 (previous step)

2 * 2 ^ i0 ≤ i1, 0 ≤ i0, 0 ≤ i1, j1 < 2, 0 ≤ j1, i1 * 2 + j1 < 6 * 2 ^ i0, 4 * 2 ^ i0 ≤ i1 * 2 + j1, 0 < i1 * 2 + j1, 0 < 2 ^ i0, ∀ j0, i. i→n(j0) < i→n(i1 * 2 + j1) → 0 ≤ i ∧ j0 < 3 * 2 ^ i ∧ 2 * 2 ^ i ≤ j0 ∧ 0 < 2 ^ i → # uint2bits(j0) = i→n(i + 3) ⊦ ~1 = i0, i1 * 2 + j1 = 1, # uint2bits(i1) + 1 = i→n(4 + i0), ¬ i1 < 2 * 2 ^ i0 ∧ ¬ 3 * 2 ^ i0 ≤ i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i * j = j * i;