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Proof step 3 from proof del-feq (previous step)

b ≠ a ∧ b ∈ s ↔ b = b ∨ b ∈ s0, ∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 = b ∨ a0 ∈ s0 ⊦ a ≠ b ∧ b ∈ s ∧ ¬ a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ s0, b ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a0. a0 ≠ a ∧ a0 ∈ s ↔ a0 = b ∨ a0 ∈ s0 ⊦ b = a

Rule argument

Additional information

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