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Proof step 19 from proof sublist-move (previous step)

m1 ≤ # x0, m ≤ m1 + 1, m ≤ 1 + # x0, ∀ n0, m0, n, m. n0 + m0 ≤ # x0 ∧ n + m ≤ m0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ a ' + firstn(m - 1, firstn(m1, x0)) = firstn(m, a ' + x0), m = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m2]

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);%firstn-firstn: ⊦ m ≤ n ∧ n ≤ # x → firstn(m, firstn(n, x)) = firstn(m, x);% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ (a ' + x).head = a; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)% ⊦ a + x = a ' + x; (s)% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m2