Definitions:-
$I$-Cauchy: S be a set and $(X,\Phi)$ is a uniform space. Let $I$ be an ideal of $S$.Define $$f:S\rightarrow X$$ $f$ is $I-Cauchy$ if for any $U\in \Phi$ there exists an $m\in S$ such that $$\{n\in S:(f(m),f(n))\notin U\}\in I$$
$I^K$-Convergent: S be a set and $(X,\Phi)$ is a uniform/topological space. Let $I$ be an ideal of $S$.Define $$f:S\rightarrow X.$$ $f$ is $I^K$ convergent to $x$ if there is an $M\in F(I)$ such that the function $$g(s)=f(s);s\in M \\=x;s\notin M$$ is $K$ convergent to $x.$ This is paraphrased as "when $f|_M$ is $K|_M$ convergent to $x$"
$I^K$-Cauchy: S be a set and $(X,\Phi)$ is a uniform space. Let $I,K$ be ideals of $S$.A map $f:S\rightarrow X$ is called $I^K$-Cauchy if $\exists$ $M\in F(I)$ such that $f|_M$ is $K|_M$-Cauchy.
And here I try to paraphrase it as follows: S be a set and $(X,\Phi)$ is a uniform space. Let $I,K$ be ideals of $S$.A map $f:S\rightarrow X$ is called $I^K$-Cauchy if the function $$g(s)=f(s);s\in M\\=(some\ fixed\ element\ say\ x_0\in X);s\notin M$$ is $K$-Cauchy.I hope this is correct.
Then here is the theorem that $f$ is $I^I$-Cauchy iff $f$ is $I$-Cauchy.The proof is given in the paper but for my clear understanding I'm writing it in my own way , I just need it correction.
proof: Let $f$ be $I^I$-Cauchy$\implies g$ is $K$-Cauchy$\implies$ for any $U\in \Phi \exists m\in S$ s.t. $\{n\in S:(g(n),g(m))\in U\}\in F(I)$.Name $$F=\{n\in S:(g(n),g(m))\in U\}.$$ So when $x,y\in F\bigcap M$ it is trivial that taking the properties of both sets $F$ and $M$, $(f(x),f(y))\in U.$ A little adjustment is required wrt $m\in S$ by using the property of uniform spaces that for every $U\in \Phi$, there exists $V\in\Phi$ s.t $V\circ V\subseteq U$.
Conversely, suppose $f$ is a $I$-Cauchy function $\implies \exists m\in S$ s.t. for any nbd $U\in \Phi$ $$\{n\in S:(f(m),f(n))\in U\}\in F(I)$$ I take this set as $M.$ Then our function $g$ that takes values of $f$ on $M$ could be $I$-Cauchy for $\{n\in S:(g(m),g(n))\in U\}\supset \{n\in S:(g(m),g(n))\in U\}=M\in F(I)\implies \{n\in S:(g(m),g(n))\in U\}\in F(I).$ The same $"m"$ that worked for $f$ also works for $g$ here.(proved)
Actually in the paper it is done with taking $M=S.$ I just tried to do with a little less trivial set as $M$ is all.
Any suggestions or corrections are appreciated. Thanks in advance.
This is Proposition $3.7(ii)$ from the paper $I^K$-Cauchy Functions by Sleziak,Toma,Das