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Proof step 2 from proof firstN-lastN-in-01 (previous step)

n < # x, a ∈ x ⊦ a ∈ firstn(# x - n, x) ↔ true, a ∈ lastn(n, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # x, a ∈ x ⊦ x = [], a ∈ firstn(# x - n, x), a ∈ lastn(n, x)

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic