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Proof step 1 from proof del-e (conclusion)

⊦ ¬ a ∈ s → ((s ++ a) -- b = ∅ ↔ s = ∅ ∧ a = b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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