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Proof step 88 from proof bitsor-bi-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 89:
bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3)) = ~1, (¬ boolvar1) ∈ bits0, bitsor(boolvar1 ' + bits0, bits3).first, boolvar ∨ boolvar0, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bitsor(boolvar1 ' + bits0, bits3) = [], bits3 = [], bits2int(bitsor(sgn(boolvar1 ' + bits0), bits3) + true ') = bits2int(bitsor(boolvar1 ' + bits0, bits3) + true '), false ∈ bitsor(boolvar1 ' + bits0, bits3), ¬ boolvar ∧ ¬ boolvar0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ false ∈ bits ∧ bits ≠ [] → bits2int(bits) = ~1;