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Proof step 4 from proof nodups-oc-2 (previous step)

n < # A, nodups(A), m < # A, n < m, n + 1 < # A ⊦ #oc(A[n], A, m) = 1

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
⊦ m < # A ∧ nodups(A) → (∀ n. (n < m → #oc(A[n], A, m) = 1) ∧ (m ≤ n ∧ n < # A → #oc(A[n], A, m) = 0))

Rule argument

spec: inst: name:nodups-oc seq: ⊦ m < # A ∧ nodups(A) → (∀ n. (n < m → #oc(A[n], A, m) = 1) ∧ ( m ≤ n ∧ n < # A → #oc(A[n], A, m) = 0)) subst:[m, A] → [m, A] flags:current, []

Additional information

[Concretesimprules(()), Extraterms((mA))]