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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: ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%unin: ⊦ ¬ a ∈ s → (a ∈ s ∪ s0 ↔ a ∈ s0);%In-empty: ⊦ ¬ a ∈ ∅; from specification infset-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic