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Proof step 4 from proof unsubdel (previous step)

a0 ∈ s, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, a0 ∈ s ∨ a0 ∈ s1 ⊦ a ∈ s1, a ∈ s, a0 ≠ a ∧ (a0 ∈ s2 ∨ a0 ∈ s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 ∈ s, a1 ∈ s, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2 ⊦ a0 ∈ s0, a0 ∈ s2, a ∈ s1, a ∈ s

Rule argument

Additional information

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