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Proof step 6 from proof bits2uint-zeros (previous step)

¬(bits ≠ [] ∧ bits.last), ∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits) = 0, true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits) = 0, bits = [], true ∈ bits, bits.last

Rule argument

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0;