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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a2 ∈ s, a1 ∈ s, a3 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a0 ∈ s ↔ a0 = b ∨ a0 = a, a1 = b, a2 = a

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ ∅ ++ b ↔ a = b; (s)% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)