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Proof step 1 from proof sgn-base-05 (conclusion)

⊦ (bits.first ↔ ¬ bits.rest.first) ∧ 1 < # bits → sgn(bits) = bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bits.first ↔ ¬ bits.rest.first ⊦ sgn(bits) = bits, # bits = 1, bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);