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Proof step 11 from proof bitsxor-empty (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ bits0 ≠ [] ∧ bits = true ' → bitsxor(bits, bits0) = bitsinvert(bits0)
Proof step 13:
¬(bits0 ≠ [] ∧ bits = true '), bitsxor(bits, bits0) = [], # bits = 1 ⊦ bits = [], bits0 = [], bits0 = true ', bits0 = false ', # bits0 = 1

Rule argument

name: bitsxor-rec-03 rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

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