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Proof step 1 from proof bits2int-011x (conclusion)

⊦ ¬ false ∈ bits → bits2int(false ' + bits) = 2 ^ n→i(# bits) - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ 2 ^ n→i(# bits) = bits2uint(bits) + 1, false ∈ bits

Rule argument

Additional information

Used simplifier rules: bits2int-pos: ⊦ bits2int(false ' + bits) = bits2uint(bits);% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;