I'm trying to figure out the Transitive Closure of T. The problem is that I cannot find anything which satisfies (x,y) (y,z) to give (x,z). Any pointers appreciated.
$$R,S,T:0..10 \iff 0..10 $$ $$T = R \circ S $$ $$ R = \{(0,8),(1,7),(2,6),(3,5),(5,3),(6,2),(7,1),(8,0)\}$$ $$S = \{(0,8),(1,7),(2,6),(3,5)\} $$ I've calculated T using: $$R \circ S = \{x \colon R, y \colon S \mid x.2 = y.1 \bullet (x.1,y.2)\}$$ Therefore: $$T = \{(5,5),(6,6),(7,7),(8,8)\}$$ Reflexive Closure: $$T \cup \{(0,0),(1,1),(2,2),(3,3),(4,4),(9,9),(10,10)$$ Symmetric Closure: $$\{(5,5),(6,6),(7,7),(8,8)\}$$ Transitive Closure: $$???$$
If there are no elements $x,y,z$ with $(x,y) \in T$ and $(y,z) \in T$, then $T$ is already transitive. This is a case of vacuous truth: The definition of transitivity $$(x,y) \in T, (y,z) \in T \rightarrow (x,z)\in T$$ is automatically satisfied if the left side can never hold.