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Proof step 46 from proof is-encoded-split-one (previous step)

# buf = m * ENCODED_NAT_SIZE, # buf2 = ENCODED_NAT_SIZE, # buf1 = ENCODED_NAT_SIZE, # ns = m + 1, m + 1 = #((ns -- n0) ++ n), is-encoded(n0, buf2), is-encoded(ns -- n0, buf), is-encoded(n, buf1), is-encoded((ns -- n0) ++ n, buf + buf1), n0 ∈ ns, n1 = n ∨ n1 ≠ n0 ∧ n1 ∈ ns, ∀ ns0, buf0, ns2, buf1, n0. # ns0 < #(ns ++ n) → # buf0 = # ns2 * ENCODED_NAT_SIZE ∧ # buf1 = ENCODED_NAT_SIZE ∧ is-encoded(n0, buf1) ∧ is-encoded(ns2, buf0) → (is-encoded(ns0, buf0 + buf1) ↔ ns0 = ns2 ++ n0) ∨ n0 ∈ ns2 ⊦ ns = ∅, n ∈ ns, n1 ≠ n0 ∧ (n1 = n ∨ n1 ∈ ns)

Proof step 46 from proof is-encoded-split-one (previous step)

User applied rule case distinction

And closed the goal!

Rule argument

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