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Proof step 18 from proof uint2bits-len (previous step)

2 ^ (i0 + 1) < 2 * 2 ^ i0, 0 ≤ 2 ^ (2 + i0), 1 < 2 ^ (2 + i0), 0 < 2 ^ (2 + i0), 2 ^ (i0 + 1) < 2 ^ (2 + i0), 0 ≤ i0, 0 < j, j < 2 ^ (2 + i0), 2 ^ (i0 + 1) ≤ j, 0 ≤ j, j < 2 * 2 ^ i0, 0 < 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) ⊦ ~1 = i0, j = 1, # uint2bits(j / 2) + 1 = i→n(3 + i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [i0 + 1, 2] rotate: F spec: int-pot inst: name: Pot-pos

Additional information

Used simplifier rules: ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ (i + 1)-1 = i;% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0);% ⊦ 1 < i ∧ 0 ≤ j → (i ^ j < i ^ j0 ↔ j < j0);% ⊦ ¬ i < i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i; Used terms: i0 + 1, 2