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Proof step 3 from proof del1 (previous step)

#oc(a, x) = m + 1 ∧ #oc(a, x) ≠ 0, a ∈ x ⊦ #oc(a, x -1l a) = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
#oc(a, x) = m + 1, a ∈ x ⊦ #oc(a, x -1l a) = m

Rule argument

Additional information

Used simplifier rules: in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;