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Proof step 9 from proof uint2bits-len (previous step)

i = i0 + 1, 2 ^ i ≤ j, j < 2 ^ (i + 1), 0 < j, 0 < 2 ^ (i + 1), 0 ≤ 2 ^ (i + 1), 1 < 2 ^ (i + 1), 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 / 2) + 1 = i→n(i + 2), j = 1, 0 = i, ¬ 2 ^ i ≤ j / 2 ∧ ¬ j / 2 < 2 ^ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
2 ^ (i0 + 1) ≤ j, j < 2 ^ (2 + i0), 0 < j, 0 ≤ i0, 2 ^ (i0 + 1) < 2 ^ (2 + i0), 0 < 2 ^ (2 + i0), 1 < 2 ^ (2 + i0), 0 ≤ 2 ^ (2 + i0), ∀ 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 / 2) + 1 = i→n(3 + i0), j = 1, ~1 = i0, ¬ 2 ^ (i0 + 1) ≤ j / 2 ∧ ¬ j / 2 < 2 ^ i0

Rule argument

Additional information

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