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Proof step 1 from proof bits2int-inj (conclusion)

⊦ bits ≠ [] ∧ bits0 ≠ [] → (bits2int(bits) = bits2int(bits0) ↔ sgn(bits) = sgn(bits0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ bits = [], bits0 = [], bits2int(bits) = bits2int(bits0) ↔ sgn(bits) = sgn(bits0)

Rule argument

Additional information