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Proof step 6 from proof bits2int-rec-03 (previous step)

¬ true ∈ bits0, false ∈ bits0 ⊦ bits2uint(bitsinvert(bits0)) + 1 = bits2uint((false ' + bitsinvert(bits0))+1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
false ∈ bits0 ⊦ bits2uint(bitsinvert(bits0)) + 1 = 2 ^ n→i(# bits0), true ∈ bits0

Rule argument

Additional information

Used simplifier rules: bits2uint-2pot: ⊦ ¬ true ∈ bits → bits2uint(true ' + bits) = 2 ^ n→i(# bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ false ' +1 = true ';% ⊦ ¬ true ∈ bits0 → (bits + bitsinvert(bits0))+1 = bits +1 + bits0;