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Proof step 1 from proof uint2bits-2pot-04 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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