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Proof step 6 from proof sublist-lenrec-01 (previous step)

n0 +1 ≤ # y, ∀ y, a. n0 ≤ # y → a ' + firstn(n0, y) = firstn(n0, a ' + y) + (a ' + y)[n0] ' ⊦ 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 7:
n0 < # y, ∀ y, a. n0 ≤ # y → a ' + firstn(n0, y) = firstn(n0, a ' + y) + (a ' + y)[n0] ' ⊦ y.head ' + firstn(n0, y.tail) = firstn(n0, y) + y[n0] ', y = []

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);%get-succ: ⊦ (a ' + x)[n + 1] = x[n];% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ (a ' + x).tail = x; (s)% ⊦ (a ' + x).head = a; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ m +1 = m + 1; (s)