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Proof step 1 from proof unall (conclusion)

⊦ s = s0 ∪ s ↔ (∀ a. a ∈ s0 → a ∈ s)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s0, a0 ∈ s0, ∀ a. a ∈ s0 → a ∈ s ⊦ a ∈ s

Rule argument

Additional information

Used simplifier rules: unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic