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Proof step 5 from proof uint2uint-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar = bits + boolvar ';