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Proof step 58 from proof bitsxor-is-orand (previous step)

bitsand(bitsinvert(bits0), bits2) = [], ∀ 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((¬ boolvar0 ∧ boolvar)', bitsand(bitsinvert(bits2), bits0) + (¬ boolvar ∧ boolvar0)'))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 59:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ []

Rule argument

name: and-notempty rotate: F subst: [bits0, bits] → [bits2, bitsinvert(bits0)]

Additional information

Used simplifier rules: ⊦ bitsinvert(bits) = [] ↔ bits = []; Used terms: bits2, bitsinvert(bits0)