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Proof step 3 from proof sgn-append-01 (previous step)

∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
false ∈ bits0 ∧ true ∈ bits0 → (∀ bits1. sgn(bits0 + bits1) = sgn(bits0) + bits1) ⊦ sgn((boolvar ' + bits0) + bits1) = sgn(boolvar ' + bits0) + bits1, ¬ false ∈ bits0 ∧ boolvar, ¬ true ∈ bits0 ∧ ¬ boolvar

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar + bits = boolvar ' + bits;