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Proof step 52 from proof one-oc (conclusion)

⊦ #oc(n, A, m0) = 1 ↔ (∃ m. m < m0 ∧ A[m] = n ∧ (∀ m1. m1 ≠ m ∧ m1 < m0 → A[m1] ≠ n))

User applied rule structural induction

And obtained the following new goals

Proof step 1:
⊦ #oc(n, A, 0) = 1 ↔ (∃ m. m < 0 ∧ A[m] = n ∧ (∀ m1. m1 ≠ m ∧ m1 < 0 → A[m1] ≠ n))

Rule argument

indvar:m0

Additional information

[Concretesimprules((Csimpgen(nat freely generated by 0, +1;)))]