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Proof step 19 from proof ubits2bits (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first)

Rule argument

seq: ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first) subst: [bits0, bits] → [false ', bits1] rotate: F spec: list inst: bitlista name: first-01

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0;% ⊦ # [] = 0;% ⊦ bitsinvert([]) = []; Used terms: false ', bits1