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Proof step 5 from proof sublist-lenrec (previous step)

m0 +1 + n < # x, ∀ n, x. m0 + n < # x → sublist(m0, n + 1, x) = sublist(m0, n, x) + x[m0 + n] ⊦ sublist(m0 +1, n + 1, x) = sublist(m0 +1, n, x) + x[m0 +1 + n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(m0 + 1) + n < # x, ∀ n0, x0. m0 + n0 < # x0 → restn(m0, x0).head ' + firstn(n0, restn(m0, x0).tail) = firstn(n0, restn(m0, x0)) + x0[m0 + n0] ' ⊦ restn(m0, x.tail).head ' + firstn(n, restn(m0, x.tail).tail) = firstn(n, restn(m0, x.tail)) + x.tail[m0 + n] ', x = []

Rule argument

Additional information

Used simplifier rules: sub-rec-01: m + n < # x ⊦ sublist(m + 1, n, x) = sublist(m, n, x.tail);%sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);%get-rec-01: ⊦ x ≠ [] → x[n + 1] = x.tail[n];% ⊦ x + a = x + a '; (s)% ⊦ ¬ m + n < m; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ # x = 0 ↔ x = []; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m +1 = m + 1; (s)