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Proof step 10 from proof ubits2bits (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ bits ≠ [] ∧ bits.last → bits2uint(bits) = bits2uint(bits.butlast) * 2 + 1
Proof step 12:
¬(bits ≠ [] ∧ bits.last), bits.first, ∀ bits0. # bits0 < # bits → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ 2 ^ n→i(# bits) = bits2uint(bits) + bits2uint(bitsinvert(bits)) + 1, bits = []
Proof step 21:
bits ≠ [] ∧ bits.last, bits.first, ∀ bits0. # bits0 < # bits → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ 2 ^ n→i(# bits) = (bits2uint(bits.butlast) * 2 + 1) + bits2uint(bitsinvert(bits)) + 1, bits = []

Rule argument

name: bits2uint-rec rotate: F subst: [bits] → [bits]

Additional information

Used terms: bits