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Proof step 5 from proof sublist-lenrec-01 (previous step)

∀ y, a. n0 ≤ # y → a ' + firstn(n0, y) = firstn(n0, a ' + y) + (a ' + y)[n0] ', n0 +1 ≤ # y ⊦ a ' + firstn(n0 +1, y) = firstn(n0 +1, a ' + y) + (a ' + y)[n0 +1] '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 < # y, ∀ y, a. n0 ≤ # y → a ' + firstn(n0, y) = firstn(n0, a ' + y) + (a ' + y)[n0] ' ⊦ y.first ' + firstn(n0, y.rest) = firstn(n0, y) + y[n0] ', y = []

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%get-succ: ⊦ (a ' + x)[n + 1] = x[n];%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (x + y) + z = x + y + z; 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%succ: ⊦ m +1 = m + 1; from specification nat-basic2