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Answer: Clarification: For sets of propositions U and V, "U entails V" is not defined. So, reformulate the question: Let U and V be propositions. Then ... Answer to this question: 1. (S1 implies S2): Assume U entails V. Then each model of U is a model of V. Thus, Mod(U) is a subset of Mod(V). 2. (S2 implies S3): Assume Mod(U) is a subset of Mod(V). Let x be a proposition such that V entails x. Then each model of V is a model of x. Now, if i is a model of U, then i is a model of V, due to our assumption. Since V entails x, i is a model of x. We have shown: each model i of U is a model of x. Therefore, U entails x. Hence, {B: V entails B} is a subset of {C: U entails C}. 3. (S3 implies S1): Assume {B: V entails B} is a subset of {C: U entails C}. Then for each proposition x, if V entails x, then U entails x. In particular, V entails V. Therefore, U entails V. Thus, equivalence of S1, S2, S3 are proved.

Answer: See the answer to an earlier instance of this question, where we answer the question when U and V are propositions. For sets of propositions U and V, there is an interpretation of entailment. It is: U entails V iff each model of all propositions in U is a model of some proposition in V. With this interpretation also, the earlier argument (language modified) proves the equivalence of S1, S2, S3. Note: alternatively use compactness.