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Proof step 6 from proof del-e (previous step)

∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ b ⊦ b ∈ s, ∀ a0. ¬ a0 ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ∈ s, a0 ∈ s, ∀ a0. ¬ a0 ∈ s ⊦ b ∈ s

Rule argument

Additional information

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