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Proof step 1 from proof del-e-02 (conclusion)

⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (∀ a0. a0 = a ∨ ¬ a0 ∈ s) ↔ (∀ a0. ¬ a0 ∈ s) ∨ (∀ a0. a0 ∈ s ↔ a0 = a)

Rule argument

Additional information

Used simplifier rules: del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b; from specification set-basic%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic