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Proof step 11 from proof del-two (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a3 ∈ s, a2 ∈ s, a1 ∈ s, a0 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a2 = b, a3 = a, a1 = a, a3 = b

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