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Proof step 2 from proof eq-01 (previous step)

⊦ s ++ a = s0 ++ a ↔ s = s0, a ∈ s0, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ (∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0)) ↔ (∀ a. a ∈ s ↔ a ∈ s0), a ∈ s, a ∈ s0

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;