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Proof step 16 from proof bits2uint-inv-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bitsinvert(bits0)) + bits2uint(bits0) + 1 ∨ bits0.first ⊦ bits = [], 2 ^ n→i(# bits) = bits2uint(bitsinvert(bits)) + bits2uint(bits.butlast) * 2 + 1, bits.last, bits.first

Proof step 16 from proof bits2uint-inv-neg (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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