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Proof step 7 from proof bits2int-rec-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ ¬ false ∈ bits ∧ bits ≠ [] → bits2uint(bits) = 2 ^ n→i(# bits) - 1

Rule argument

name: bits2uint-ones rotate: F subst: [bits] → [bitsinvert(bits0)]

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0;% ⊦ # [] = 0;% ⊦ bitsinvert([]) = [];% ⊦ false ∈ bitsinvert(bits) ↔ true ∈ bits;% ⊦ # bitsinvert(bits) = # bits;% ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ bitsinvert(bits) = [] ↔ bits = []; Used terms: bitsinvert(bits0)