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Proof step 1 from proof del-two (conclusion)

s -- a -- b = ∅ ⊦ s = ∅ ∨ s = ∅ ++ a ∨ s = ∅ ++ b ∨ s = ∅ ++ a ++ b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a3 ∈ s ↔ a3 ∈ ∅ ++ a ++ b, a2 ∈ s ↔ a2 = b, a1 ∈ s ↔ a1 = a

Rule argument

Additional information

Used simplifier rules: del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b; from specification set-basic%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic