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For any hypergraph H = V, E and any sets A, B1 , . . , Bn ⊆ E, if H 1≤i≤n [Bi ] → [A], then there is a finite protocol P over H such that P [A] and P [Bi ] for all i ≤ n. Proof. Induction on the size of V . Case 1. If |A| ≤ 1, then, by the Small Set axiom, H 1≤i≤n [Bi ] → [A], which is a contradiction. [A]. Hence, H Case 2. Suppose that the edges of hypergraph H can be divided into two nontrivial disconnected sets X and Y . Thus, the empty set is a gateway between A ∩ X and A ∩ Y . By the Gateway axiom, H [A ∩ X] → ([A ∩ Y ] → [A]).

He arranged them in a circle and told them that he would put a white or a black spot on their foreheads and that one of the three spots would certainly be white. The three wise men could see and hear each other but, of course, they could not see their faces reﬂected anywhere. The king, then, asked to each of them to ﬁnd out the color of his own spot. After a while, the wisest correctly answered that his spot was white. We employ a 4-dimensional logic combining the modalities ✷a , ✷b , and ✷c , for encoding the individual knowledge of the three wise men, and a box operator ✷fool , for encoding the knowledge that is common to all of them.

H 1≤i≤n By the laws of propositional logic, H [Bi0 ] → ( [Bi ] → [A \ In(VY )]). 1≤i≤n Since Out(VY ) ⊆ Bi0 , by Theorem 1, [Out(VY )] → ( H [Bi ] → [A \ In(VY )]). 1≤i≤n Again by Theorem 1, H [Out(VY )] → ( [Bi \ In(VY )] → [A \ In(VY )]). 1≤i≤n Let H be the result of the truncation of set VY from hypergraph H. By the Truncation rule, H [Bi \ In(VY )] → [A \ In(VY )]. 1≤i≤n [A \ By the Induction Hypothesis, there is a protocol P on H such that P In(VY )] and P [Bi \ In(VY )] for any i ≤ n. Therefore, by Theorem 14, there is a protocol P on H such that P [A] and P [Bi ] for any i ≤ n.