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Proof step 16 from proof is-encoded-split-one (
previous step
)
⊦ ns ≠ ∅ → (is-encoded(ns, buf) ↔ (∃ n, buf0, buf1. n ∈ ns ∧ buf = buf0 + buf1 ∧ is-encoded(n, buf0) ∧ is-encoded(ns -- n, buf1)))
Lemma
is-encoded-rec