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Proof step 1 from proof in-at (conclusion)

⊦ n < # ar ∧ ar[n] = a → (¬ a ∈ ar ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < # ar ⊦ # ar = 0, ∃ n0. n0 < # ar ∧ ar[n0] = ar[n]

Rule argument

Additional information

Used simplifier rules: in: ⊦ a ∈ ar ↔ (∃ n. n < # ar ∧ ar[n] = a);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1