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Proof step 4 from proof firstn-restn-in (previous step)

n0 +1 ≤ # x, ∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)) ⊦ a ∈ x ↔ a ∈ firstn(n0 +1, x) ∨ a ∈ restn(n0 +1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 < # x, ∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)) ⊦ a ∈ x ↔ a = x.head ∨ a ∈ firstn(n0, x.tail) ∨ a ∈ restn(n0, x.tail), x = []

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);% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%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)