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Proof step 9 from proof del-e-02 (previous step)

(∀ a0. ¬ a0 ∈ s) ∨ (∀ a0. a0 ∈ s ↔ a0 = a) ⊦ ∀ a0. a0 = a ∨ ¬ a0 ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a0 ∈ s, a1 ∈ s, (∀ a0. ¬ a0 ∈ s) ∨ (∀ a0. a0 ∈ s ↔ a0 = a) ⊦ a0 = a

Rule argument

Additional information

Used simplifier rules: In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic