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

1 ≤ # bits, ∀ bits, bits1. # bits1 < 1 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ bits = [], bitsxor(bits, false ') = bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ bits ≠ [] ∧ bits0 = false ' → bitsxor(bits, bits0) = bits

Rule argument

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

Additional information

Used terms: false ', bits