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Proof step 3 from proof no-oc (previous step)

#oc(A[m0], A, m) = 0, m0 ≤ m, ∀ n, A. (#oc(n, A, m) = 0 ↔ (∀ m1. m1 < m → A[m1] ≠ n)) ⊦ A[m] = A[m0]

User applied rule all left

And obtained the following new goals

Proof step 2:
#oc(A[m0], A, m) = 0, m0 ≤ m, #oc(A[m0], A, m) = 0 ↔ (∀ m1. m1 < m → A[m1] ≠ A[m0]), ∀ n, A. (#oc(n, A, m) = 0 ↔ (∀ m1. m1 < m → A[m1] ≠ n)) ⊦ A[m] = A[m0]

Rule argument

pos:left 3 terms:(A[m0]A) abort:F discard:F computed:T

Additional information

[Concretesimprules(()), Extraterms((A[m0]A))]