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Proof step 3 from proof firstn-restn-in (previous step)

∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)), n0 +1 ≤ # 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 4:
n0 < # x, ∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)) ⊦ a ∈ x ↔ a = x.first ∨ a ∈ firstn(n0, x.rest) ∨ a ∈ restn(n0, x.rest), x = []

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%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%a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); 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