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Proof step 34 from proof is-encoded-split-one (previous step)

# buf1 = ENCODED_NAT_SIZE, # buf2 = ENCODED_NAT_SIZE, # buf = (# ns - 1) * ENCODED_NAT_SIZE, ENCODED_NAT_SIZE + # buf = # ns * ENCODED_NAT_SIZE, is-encoded(n, buf1), is-encoded(ns -- n0, buf), is-encoded(n0, buf2), n0 ∈ 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 = 0, ns = ∅, n ∈ ns, ∃ n0, buf4, buf3. (n0 = n ∨ n0 ∈ ns) ∧ (buf2 + buf) + buf1 = buf4 + buf3 ∧ is-encoded(n0, buf4) ∧ is-encoded((ns ++ n) -- n0, buf3)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 35:
# buf = m * ENCODED_NAT_SIZE, # buf2 = ENCODED_NAT_SIZE, # buf1 = ENCODED_NAT_SIZE, # ns = m + 1, ENCODED_NAT_SIZE + # buf = # ns * ENCODED_NAT_SIZE, is-encoded(n0, buf2), is-encoded(ns -- n0, buf), is-encoded(n, buf1), n0 ∈ 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 = 0, ns = ∅, n ∈ ns, ∃ n0, buf4, buf3. (n0 = n ∨ n0 ∈ ns) ∧ (buf2 + buf) + buf1 = buf4 + buf3 ∧ is-encoded(n0, buf4) ∧ is-encoded((ns ++ n) -- n0, buf3)

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# ns, m]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # ns, m