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Proof step 13 from proof is-encoded-unique (previous step)

# ns0 = m + 1, # buf = m * ENCODED_NAT_SIZE, # buf2 = ENCODED_NAT_SIZE, m + 1 = # ns1, is-encoded(n0, buf2), is-encoded(ns0 -- n0, buf), is-encoded(ns1 -- n0, buf), n0 ∈ ns1, n0 ∈ ns0, ∀ ns, buf, ns1. # ns < m + 1 → # buf = # ns1 * ENCODED_NAT_SIZE ∧ # buf = # ns * ENCODED_NAT_SIZE ∧ is-encoded(ns, buf) ∧ is-encoded(ns1, buf) → ns = ns1 ⊦ ns0 = ∅, ns0 = ns1, ns1 = ∅

User applied rule apply induction

And closed the goal!

Rule argument

subst: [ns, buf, ns1] → [ns0 -- n0, buf, ns1 -- n0]

Additional information

Used simplifier rules: ⊦ n ∈ ns ∧ n ∈ ns0 → (ns -- n = ns0 -- n ↔ ns = ns0); (s)% ⊦ n ∈ ns → #(ns -- n) = # ns - 1; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ m < n → ¬ m + m0 < n; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) Used terms: ns0 -- n0, buf, ns1 -- n0