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Proof step 1 from proof sublist-lenrec-01 (conclusion)

⊦ n < # x → sublist(0, n + 1, x) = sublist(0, n, x) + x[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < # x ⊦ x.first ' + firstn(n, x.rest) = firstn(n, x) + x[n] ', x = []

Rule argument

Additional information

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