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

false ∈ bits, true ∈ bits ⊦ sgn(bits + bits0) = sgn(bits) + bits0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ bits1. false ∈ bits0 ∧ true ∈ bits0 → sgn(bits0 + bits1) = sgn(bits0) + bits1 ⊦ false ∈ boolvar + bits0 ∧ true ∈ boolvar + bits0 → sgn((boolvar + bits0) + bits1) = sgn(boolvar + bits0) + bits1

Rule argument

indvar: bits

Additional information

Used simplifier rules: sgn-empty: ⊦ sgn([]) = [];%bitlist freely generated by [], +;;% ⊦ [] + bits = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ boolvar ∈ [];