I) t ∈ α(A ∪ B) ⇒ there is an s ∈ A ∪ B such that α(s) = t ⇒ there is an s ∈ A or s ∈ B such that α(s) = t ⇒ t ∈ α(A) or t ∈ α(B). Thus α(A ∪ B) ⊂ α(A) ∪ α(B). t ∈ α(A) ∪ α(B) ⇒ there is an s ∈ A or s ∈ B such that α(s) = t ⇒ there is an s ∈ A ∪ B such that α(s) = t ⇒ t ∈ α(A ∪ B). Thus α(A) ∪ α(B) ⊂ α(A ∪ B).

II) t ∈ α(A ∩ B) ⇒ there is an s ∈ A ∩ B such that α(s) = t ⇒ there is an s ∈ A and s ∈ B such that α(s) = t ⇒ t ∈ α(A) and t ∈ α(B). Thus α(A ∩ B) ⊂ α(A) ∩ α(B).

III) α = { (s_{1}, t_{1}), (s_{2}, t_{0}), (s_{3}, t_{0}) }. A ⊂ S = { s_{1}, s_{2} }. B ⊂ S = { s_{1}, s_{3} }. α(A ∩ B) = { t_{1} }. α(A) ∩ α(B) = { t_{1}, t_{0} }.

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