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Proof step 14 from proof bitsxor-empty (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ 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 '