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Proof step 56 from proof bitsxor-is-orand (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 4

Additional information