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Proof step 2 from proof sgn-last-04 (previous step)

(¬ boolvar) ∈ bits ⊦ sgn(bits + boolvar ') = sgn(bits) + boolvar '

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ (¬ boolvar0) ∈ boolvar + bits0 → sgn((boolvar + bits0) + boolvar0 ') = sgn(boolvar + bits0) + boolvar0 '

Rule argument

indvar: bits

Additional information

Used simplifier rules: sgn-one: ⊦ sgn(boolvar ') = boolvar ';%sgn-empty: ⊦ sgn([]) = [];%bitlist freely generated by [], +;;% ⊦ [] + bits = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;