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Proof step 13 from proof bits2uint-inv-neg (previous step)

bits1 = [] ∨ 2 ^ n→i(# bits1) = bits2uint(bitsinvert(bits1)) + bits2uint(bits1) + 1 ∨ bits1.first, ∀ bits0. # bits0 < #(bits1 + true ') → true → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bitsinvert(bits0)) + bits2uint(bits0) + 1 ∨ bits0.first ⊦ 2 ^ (n→i(# bits1) + 1) = 2 * ((bits2uint(bitsinvert(bits1)) + bits2uint(bits1)) + 1), (bits1 + true ').first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
2 ^ n→i(# bits1) = bits2uint(bitsinvert(bits1)) + bits2uint(bits1) + 1, ∀ bits0. # bits0 < #(bits1 + true ') → true → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bitsinvert(bits0)) + bits2uint(bits0) + 1 ∨ bits0.first ⊦ 2 ^ (n→i(# bits1) + 1) = 2 * ((bits2uint(bitsinvert(bits1)) + bits2uint(bits1)) + 1), bits1 = [], bits1.first

Rule argument

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ bitsinvert([]) = [];% ⊦ # [] = 0;