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Proof step 3 from proof sgn-zeros (previous step)

∀ bits1. sgn(bits0 + false ' + bits1) = sgn(false ' + bits1) ∨ true ∈ bits0 ⊦ sgn((boolvar + bits0) + false ' + bits1) = sgn(false ' + bits1) ∨ true ∈ boolvar + bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ bits. sgn(bits0 + false ' + bits) = sgn(false ' + bits) ⊦ sgn((false ' + bits0) + false ' + bits1) = sgn(false ' + bits1), true ∈ bits0

Rule argument

Additional information

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