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Proof step 1 from proof uint2bits-2pot-05 (conclusion)

⊦ ¬ false ∈ bits ∧ 0 ≤ i → (uint2bits(i) = boolvar ' + bits ↔ ¬ boolvar ∧ i = 2 ^ n→i(# bits) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i ⊦ uint2bits(i) = boolvar ' + bits ↔ 2 ^ n→i(# bits) = i + 1 ∧ ¬ boolvar, false ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;