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

∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits) = 0, true ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ bits ≠ [] ∧ bits.last → bits2uint(bits) = bits2uint(bits.butlast) * 2 + 1
Proof step 6:
¬(bits ≠ [] ∧ bits.last), ∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits) = 0, true ∈ bits
Proof step 10:
bits ≠ [] ∧ bits.last, ∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits.butlast) * 2 + 1 = 0, true ∈ bits

Rule argument

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

Additional information

Used terms: bits