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Proof step 4 from proof sublist-lenrec (previous step)

∀ n, x. m0 + n < # x → sublist(m0, n + 1, x) = sublist(m0, n, x) + x[m0 + n] ⊦ m0 +1 + n < # x → 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 5:
(m0 + 1) + n < # x, ∀ n0, x0. m0 + n0 < # x0 → restn(m0, x0).first ' + firstn(n0, restn(m0, x0).rest) = firstn(n0, restn(m0, x0)) + x0[m0 + n0] ' ⊦ x = [], restn(m0, x.rest).first ' + firstn(n, restn(m0, x.rest).rest) = firstn(n, restn(m0, x.rest)) + x.rest[m0 + n] '

Rule argument

Additional information

Used simplifier rules: sub-rec-01: m + n < # x ⊦ sublist(m + 1, n, x) = sublist(m, n, x.rest);%sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%get-rec-01: ⊦ x ≠ [] → x[n + 1] = x.rest[n];%Last: ⊦ x + a = x + a '; from specification list%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ a + x = a ' + x; from specification list%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2