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Proof step 5 from proof unsubsub (previous step)

∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, a ∈ s ∨ a ∈ s1 ⊦ a ∈ s, a ∈ s0, a ∈ s2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ s1, a0 ∈ s1, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s2, a ∈ s0, a ∈ s

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