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Proof step 1 from proof lastn-firstn-in (conclusion)

⊦ lastn(n, x) = y ∧ ¬ a ∈ y ∧ a ∈ x ∧ n ≤ # x → a ∈ firstn(# x - n, x)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: firstN-lastN-in: ⊦ n ≤ # x ∧ a ∈ x ∧ ¬ a ∈ lastn(n, x) → (¬ a ∈ firstn(# x - n, x) ↔ false);