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Proof step 31 from proof bitsxor-is-orand (previous step)

∀ bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ # bits0 = 1, bits0 = false ', bits0 = true ', bits2int(bitsxor(false ', bits0)) = bits2int(bits0), bits0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 32:
⊦ bits0 ≠ [] ∧ bits = false ' → bitsxor(bits, bits0) = bits0

Rule argument

name: bitsxor-rec-04 rotate: F subst: [bits0, bits] → [bits0, false ']

Additional information

Used terms: bits0, false '