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Proof step 2 from proof firstn-restn-in (previous step)

n ≤ # x ⊦ a ∈ x ↔ a ∈ firstn(n, x) ∨ a ∈ restn(n, x)

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ 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)

Rule argument

indvar: n

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%restN-zero: ⊦ restn(0, x) = x;%nat freely generated by 0, +1;;%in-empty: ⊦ ¬ a ∈ []; from specification list