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Proof step 41 from proof bitsor-bi-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bitsor(boolvar ', bits3 + boolvar0 ')) = bits2int(bitsor(boolvar ' + bits0, bits3) + (boolvar ∨ boolvar0)'), bits3 = [], (¬ boolvar) ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;