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Proof step 22 from proof ubits2bits (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 23:
bits = bits1 + boolvar ', boolvar, bits.first, ∀ bits0. # bits0 < # bits → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ bits = [], 2 ^ n→i(# bits) = 2 * (bits2uint(bits1) + 1) + bits2uint(bitsinvert(bits))

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits, boolvar, bits1] rotate: T spec: list-last inst: bitlista name: last-elim

Additional information

Used simplifier rules: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar '; (elim) Used terms: bits, boolvar, bits1