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Proof step 68 from proof bitsxor-is-orand (previous step)

bits0 = [] ∨ bits2 = [] ∨ bits2int(bitsxor(bits0, bits2)) = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0))), ∀ bits, bits1. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bitsand(bitsinvert(bits0), bits2) = [], bitsand(bitsinvert(bits2), bits0) = [], bits0 = [], bits2 = [], bits2int(bitsxor(bits0, bits2) + (¬(boolvar0 ↔ boolvar))') = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0)) + (¬ boolvar0 ∧ boolvar ∨ ¬ boolvar ∧ boolvar0)')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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