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Proof step 1 from proof in-get (conclusion)

⊦ a ∈ x ↔ (∃ n. n < # x ∧ x[n] = a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ x ⊦ ∃ n. n < # x ∧ x[n] = a

Rule argument

Additional information

Used simplifier rules: get-in: ⊦ m < # x → (¬ x[m] ∈ x ↔ false);