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Proof step 27 from proof del-two (previous step)

a3 ∈ s, a4 ∈ s, a0 ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦ a3 = a, a2 ∈ s ↔ a2 = b, a3 ∈ s ↔ a3 ∈ ∅ ++ a ++ b, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
a3 ∈ s, a4 ∈ s, a0 ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦ a2 ∈ s ↔ a2 = b, a3 = b, a ∈ s

Rule argument

Additional information

Used simplifier rules: in-04: ⊦ a ∈ ∅ ++ b ↔ a = b; from specification set-basic%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic