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Proof step 3 from proof uint2bits-len-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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) = i→n(i + 3)

Proof step 3 from proof uint2bits-len-01 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information