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Proof step 143 from proof bitsor-bi-sgn (previous step)

¬(boolvar ∨ boolvar0) ∧ bitsor(boolvar1 ' + bits0, bits3) ≠ [], bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3)) = bits2int(bitsor(boolvar1 ' + bits0, bits3)), (¬ boolvar1) ∈ bits0, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits3 = [], bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3) + (boolvar ∨ boolvar0)') = bits2int(bitsor(boolvar1 ' + bits0, bits3)) * 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information