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Proof step 5 from proof uint2bits-second (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → uint2bits(i0).rest.first ⊦ i = 1, (uint2bits(i / 2) + (i % 2 = 1)').rest.first

Rule argument

Additional information

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