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Proof step 1 from proof int-eq (conclusion)

⊦ s = s ∩ s0 ↔ s ⊆ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%int-rax: ⊦ a ∈ s1 ∩ s2 ↔ ¬ ¬(a ∈ s1 ∧ a ∈ s2);%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%subdef: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2);%Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%In-empty: ⊦ ¬ a ∈ ∅; from specification infset-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic