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Proof step 6 from proof uint2bits-first (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ i, 0 < i, (uint2bits(i / 2) + (i % 2 = 1)').first, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ uint2bits(i0).first → false ⊦ i = 1

Rule argument

Additional information

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