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Proof step 84 from proof bitsxor-is-orand (previous step)

bits2int([]) = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0))), bitsxor(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 = [], bitsand(bitsinvert(bits2), bits0) = [], bitsand(bitsinvert(bits0), bits2) = [], bits2int((¬(boolvar0 ↔ boolvar))') = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0)) + (¬ boolvar0 ∧ boolvar ∨ ¬ boolvar ∧ boolvar0)')

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 85:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsxor(bits, bits0) ≠ []

Rule argument

name: bitsxor-empty rotate: F subst: [bits0, bits] → [bits2, bits0]

Additional information

Used simplifier rules: bitsxor-empty: ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsxor(bits, bits0) ≠ []; Used terms: bits2, bits0