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Proof step 8 from proof bitsxor-empty (previous step)

bitsxor(bits, bits0) = [] ⊦ bits = [], bits0 = [], bits0 = true ', # bits ≠ 1 ∧ # bits0 ≠ 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ bits ≠ [] ∧ bits0 = false ' → bitsxor(bits, bits0) = bits
Proof step 10:
¬(bits ≠ [] ∧ bits0 = false '), bitsxor(bits, bits0) = [] ⊦ bits = [], bits0 = [], bits0 = true ', # bits ≠ 1 ∧ # bits0 ≠ 1

Rule argument

name: bitsxor-rec-02 rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

Used terms: bits0, bits