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Proof step 35 from proof sgn-last-04 (previous step)

∀ boolvar. ¬ boolvar ∨ false ∈ bits1 → sgn((true ' + bits1) + boolvar ') = sgn(true ' + bits1) + boolvar ' ⊦ ¬ (¬ boolvar0) ∈ bits1, sgn(true ' + bits1 + boolvar0 ') = sgn(true ' + bits1) + boolvar0 '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
(¬ boolvar0) ∈ bits1, ∀ boolvar. ¬ boolvar ∨ false ∈ bits1 → sgn((true ' + bits1) + boolvar ') = sgn(true ' + bits1) + boolvar ' ⊦ sgn(true ' + bits1 + boolvar0 ') = sgn(true ' + bits1) + boolvar0 '

Proof step 35 from proof sgn-last-04 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information