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Proof step 6 from proof subseteq-inter-empty-empty (previous step)

∀ a. a ∈ s1 ∩ s2 ↔ a ∈ ∅, ∀ a. a ∈ s0 → a ∈ s1 ⊦ #(s0 ∩ s2) = 0, s0 ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a. a ∈ s0 → a ∈ s1, ∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s2 ⊦ #(s0 ∩ s2) = 0, s0 ∩ s2 = ∅

Rule argument

Additional information

Used simplifier rules: int-rax: ⊦ a ∈ s1 ∩ s2 ↔ ¬ ¬(a ∈ s1 ∧ a ∈ s2);%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic