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Proof step 23 from proof del-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
perm(y, x), a0 ∈ y, ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y) ⊦ #oc(a0, x) = #oc(a0, y), a0 = a

Rule argument

Additional information

Used simplifier rules: del1-01: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x);%notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%in: ⊦ a ∈ y ∧ ¬ a ∈ x → ¬ perm(x, y);%del1-in: ⊦ a ≠ b → (a ∈ x -1l b ↔ a ∈ x); from specification list-del