Transitive closures and inductive reasoning [solved] - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T08:14:06Zhttp://mathoverflow.net/feeds/question/109066http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/109066/transitive-closures-and-inductive-reasoning-solvedTransitive closures and inductive reasoning [solved]Helmut Brandl2012-10-07T14:28:20Z2012-10-07T22:53:23Z
<p>Let's say that r is an endorelation over A (i.e. $r$ is a subset of $A \times A$), $\bar{r}$ is the transitive closure of r (i.e. the least set containing r and being transitive).</p>
<p>Furthermore $r$ has the property that for all $x,y$ such that $(x,y) \in r$ and $x$ has a certain property $p(x)$ implies that $y$ has the property as well ($p(y)$ is valid).</p>
<p>I am trying to proof that the transitive closure $\bar{r}$ of r has this property as well.</p>
<p>I.e. I am trying to proof</p>
<p>\begin{equation}
(\forall x,y: (x,y) \in r \land p(x) \Rightarrow p(y)) \Rightarrow (\forall x,y: (x,y) \in \bar{r} \land p(x) \Rightarrow p(y))
\end{equation}</p>
<p>But I cannot find a proof of this assertion.</p>
<p>For me the above claimed assertion is obvious so it should have a proof. Can anybody help me to find some or to find counterexample which demonstrates that the above assertion is not valid in general.</p>
<p>It is easy to prove that $r$ and its transitive closure have the same domain and the same range. Furthermore I can prove that any domain/range restriction of $r$ results in the same domain/range restriction of its transtitive closure.</p>