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Proof step 4 from proof sgn-last (previous step)

¬ boolvar1, (¬ boolvar1) ∈ bits0, ∀ boolvar, boolvar0. (¬ boolvar) ∈ bits0 → sgn(boolvar ' + bits0 + boolvar0 ') = sgn(boolvar ' + bits0) + boolvar0 ' ⊦ boolvar, sgn(boolvar1 ' + (boolvar ' + bits0) + boolvar0 ') = sgn(boolvar1 ' + boolvar ' + bits0) + boolvar0 '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
true ∈ bits0, ∀ boolvar, boolvar0. (¬ boolvar) ∈ bits0 → sgn(boolvar ' + bits0 + boolvar0 ') = sgn(boolvar ' + bits0) + boolvar0 ' ⊦ sgn(false ' + bits0 + boolvar0 ') = sgn(false ' + bits0) + boolvar0 '

Rule argument

Additional information

Used simplifier rules: sgn-rec: ⊦ sgn(boolvar ' + boolvar ' + bits) = sgn(boolvar ' + bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;