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Proof step 2 from proof del-two (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a3 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a2 ∈ s ↔ a2 = a, a1 ∈ s ↔ a1 = b, a0 ∈ s ↔ a0 ∈ ∅ ++ a ++ b

Rule argument

Additional information

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