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Proof step 1 from proof del-eq (conclusion)

⊦ a ∈ s ∧ a ∈ s0 → (s -- a = s0 -- a ↔ s = s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s ⊦ (∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0)) ↔ (∀ a. a ∈ s ↔ a ∈ s0)

Rule argument

Additional information

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