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Proof step 21 from proof sgn-append-03 (previous step)

∀ bits1, boolvar. sgn(bits0 + bits1) = sgn((¬ boolvar)' + bits1) ∨ boolvar ∈ bits0, bits0 + bits1 = boolvar1 ' + bits ⊦ bits0 + bits1 = [], boolvar0 ↔ boolvar1, sgn(bits0 + bits1) = sgn((¬ boolvar0)' + bits1), boolvar0 ∈ bits0, ¬ boolvar1 ∈ bits0 ∧ ¬ boolvar1 ∈ bits1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: bits1, boolvar0

Additional information

Used terms: bits1, boolvar0