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Proof step 6 from proof ocs-del1 (previous step)

∀ a. a ∈ x0 → #oc(a, x0) = 1 ∨ a ∈ x0 -1l a, a = a0 ∨ a ∈ x0 ⊦ a = a0, a ∈ (a0 ' + x0) -1l a, ¬ a ∈ x0 ∧ #oc(a, a0 ') = 1, a ≠ a0 ∧ #oc(a, x0) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ∈ x0, ∀ a. a ∈ x0 → #oc(a, x0) = 1 ∨ a ∈ x0 -1l a ⊦ #oc(a, x0) = 1, a = a0, a ∈ x0 -1l a

Rule argument

Additional information

Used simplifier rules: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); (s)