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Proof step 3 from proof firstrestex (previous step)

n0 +1 < # x, ∀ x. n0 < # x → x = firstn(n0, x) + restn(n0, x) ∨ x = [] ⊦ x = [], x = firstn(n0 +1, x) + restn(n0 +1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 + 1 < # x, ∀ x. n0 < # x → x = firstn(n0, x) + restn(n0, x) ∨ x = [] ⊦ x = [], x = (x.head ' + firstn(n0, x.tail)) + restn(n0, x.tail)

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);%restN-rec: ⊦ restn(n + 1, x) = restn(n, x.tail);%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)