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Proof step 35 from proof bitsor-bi-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits3 = [], bits2int(bitsor(sgn(boolvar1 ' + boolvar '), bits3 + boolvar0 ')) = bits2int(bitsor(boolvar1 ' + bits0, bits3) + (boolvar ∨ boolvar0)'), (¬ boolvar1) ∈ bits0

Rule argument

Additional information

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