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Proof step 4 from proof no-oc (previous step)

#oc(n, A, m) + (A[m] = n ⊃ 1;0) = 0, ∀ n, A. (#oc(n, A, m) = 0 ↔ (∀ m1. m1 < m → A[m1] ≠ n)) ⊦ ∀ m0. ¬ m < m0 → A[m0] ≠ n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
#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]

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(nat,,lels, ⊦ ¬ m < n ↔ n ≤ m) Csimpnamedseq(nat-basic2,,c, ⊦ m + n = n + m) Csimpnamedseq(nat-basic2,,a, ⊦ (m + n0) + n1 = m + n0 + n1) Csimpnamedseq(nat-basic2,,z-01, ⊦ m + n = 0 ↔ m = 0 ∧ n = 0)))]