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Proof step 6 from proof uint2uint-pos (previous step)

0 < i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → bits2uint(uint2bits(i0)) = i0 ⊦ i = 1, bits2uint(uint2bits(i / 2) + (i % 2 = 1)') = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ boolvar → bits2uint(bits + boolvar ') = bits2uint(bits) * 2 + 1
Proof step 8:
i % 2 ≠ 1, 0 < i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → bits2uint(uint2bits(i0)) = i0 ⊦ i = 1, bits2uint(uint2bits(i / 2) + (i % 2 = 1)') = i
Proof step 10:
i % 2 = 1, 0 < i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → bits2uint(uint2bits(i0)) = i0 ⊦ i = 1, bits2uint(uint2bits(i / 2)) * 2 + 1 = i

Rule argument

name: bits2uint-rec-02 rotate: F subst: [boolvar, bits] → [i % 2 = 1, uint2bits(i / 2)]

Additional information

Used terms: i % 2 = 1, uint2bits(i / 2)