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Proof step 4 from proof bits2uint-ones (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ bits ≠ [] ↔ (∃ bits0, boolvar. bits = bits0 + boolvar ')
Proof step 6:
∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + 1 ∨ false ∈ bits0 ⊦ 2 ^ n→i(# bits) = bits2uint(bits) + 1, ∀ bits0, boolvar. bits ≠ bits0 + boolvar ', false ∈ bits

Rule argument

seq: ⊦ bits ≠ [] ↔ (∃ bits0, boolvar. bits = bits0 + boolvar ') subst: [bits] → [bits] rotate: F spec: list inst: bitlista name: lastdiv-rew

Additional information

Used terms: bits