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Proof step 6 from proof del-05 (previous step)

perm(x -1l a, x), a ∈ x, a ∈ x ∨ a ∈ x -1l a → #oc(a, x) = #oc(a, x -1l a), ∀ a0. a0 ∈ x ∨ a0 ∈ x -1l a → #oc(a0, x) = #oc(a0, x -1l a) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
#oc(a, x) = #oc(a, x) - 1, perm(x -1l a, x), a ∈ x, ∀ a0. a0 ∈ x ∨ a0 ∈ x -1l a → #oc(a0, x) = #oc(a0, x -1l a) ⊦

Rule argument

Additional information

Used simplifier rules: del1: ⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1;% ⊦ n -1 = n - 1; (s)