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Proof step 60 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))) ⊦ bitsand(bitsinvert(bits2), bits0) ≠ [], bits0 = [], bits2 = [], bits2int(bitsxor(bits0 + boolvar0 ', bits2 + boolvar ')) = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2) + (¬ boolvar0 ∧ boolvar)', bitsand(bitsinvert(bits2), bits0) + (¬ boolvar ∧ boolvar0)'))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 61:
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))) ⊦ bits2 = [], bits0 = [], bits2int(bitsxor(bits0 + boolvar0 ', bits2 + boolvar ')) = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2) + (¬ boolvar0 ∧ boolvar)', (¬ boolvar ∧ boolvar0)'))

Rule argument

Additional information

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