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Proof step 70 from proof bitsor-bi-sgn (previous step)

bits2 = boolvar1 + bits0, (¬ boolvar1) ∈ bits0, ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2 = [], bits3 = [], bits2int(bitsor(sgn(bits2) + boolvar ', bits3 + boolvar0 ')) = bits2int(bitsor(bits2, bits3) + (boolvar ∨ boolvar0)')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 71:
(¬ 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) + boolvar ', bits3 + boolvar0 ')) = bits2int(bitsor(boolvar1 ' + bits0, bits3) + (boolvar ∨ boolvar0)')

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;