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Proof step 3 from proof union-subset-03 (previous step)

a ∈ s, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s3 → a ∈ s4 ⊦ a ∈ s0, a ∈ s2, a ∈ s4

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ s, a0 ∈ s, ∀ a. a ∈ s3 → a ∈ s4, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s4, a ∈ s2, a ∈ s0

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