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Proof step 12 from proof no-oc (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 4:
#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

Rule argument

pos:right 1

Additional information

[Concretesimprules(())]