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Proof step 2 from proof sgn-in-01 (previous step)

⊦ bits = [], boolvar ∈ sgn(bits) ↔ boolvar ∈ bits

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ boolvar. bits0 = [] ∨ (boolvar ∈ sgn(bits0) ↔ boolvar ∈ bits0) ⊦ boolvar + bits0 = [] ∨ (boolvar0 ∈ sgn(boolvar + bits0) ↔ boolvar0 ∈ boolvar + bits0)

Rule argument

indvar: bits

Additional information

Used simplifier rules: bitlist freely generated by [], +;;