inf and min in partial order Assume $P$ is a partial order. Does it holds that $\inf (\inf(a,b), \inf(c,d)) = \inf(a,b,c,d) $? My guess is that it holds. But I don't know how to check it. Also what if instead of $\inf$ I use $\min$? It seems to be the same. Thank you for help!!
 A: By definition, if $\inf(a,b,c,d)$ exists, then it is less than or equal to each of $a,b,c,d$. Since it is less than or equal to $a,b$, then it is less than or equal to $\inf(a,b)$ (if that exists) by definition of infimum, and similarly is less than or equal to $\inf(c,d)$ (if that exists). Hence, if both sides of the proposed equation are defined, then $$\inf\bigl(\inf(a,b),\inf(c,d)\bigr)\geq\inf(a,b,c,d),$$ again by definition of infimum. On the other hand, if $\inf\bigl(\inf(a,b),\inf(c,d)\bigr)$ is defined, then it, too, is less than or equal to each of $a,b,c,d$, and so once again  by definition of infimum, we have that if both sides of the proposed equation are defined, then $$\inf\bigl(\inf(a,b),\inf(c,d)\bigr)\leq\inf(a,b,c,d).$$ By antisymmetry, the identity holds.
Similar arguments can be used with $\min$ in place of $\inf$. It's worth noting that, in a general partial order $P$, with arbitrarily chosen elements $a,b,c,d$, there may be no such thing as $\inf(a,b)$ or $\inf(c,d)$, and even if those exist, there may be no such thing as $\inf\bigl(\inf(a,b),\inf(c,d)\bigr)$. It's even less likely that they all exist when we use $\min$ instead of $\inf$, since $\min$ requires comparability to make any sense, and that needn't happen in partial orders.
A: Let $a,b,c,d\in P$.
Suppose $\inf (\inf(a,b), \inf(c,d)), \inf (a,b), \inf (c,d)$ and $\inf(a,b,c,d)$ exist.
By definition of $\inf$ we have 


*

*$\inf (a,b)\leq a \wedge \inf (a,b)\leq b$ 

*$\inf (c,d)\leq c \wedge \inf (c,d) \leq d$. 

*$\inf(\inf(a,b),\inf(c,d))\leq \inf(a,b) \wedge \inf(\inf(a,b),\inf(c,d))\leq \inf(c,d)$


Therefore by the transivity of the the relation $\leq$, it follows that $\inf(\inf(a,b),\inf(c,d))\leq x$, for all $x\in \{a,b,c,d\}$. This means that $\inf(\inf(a,b),\inf(c,d))$ is a lowerbound for $\{a,b,c,d\}$ and therefore $\color{blue}{\inf(\inf(a,b),\inf(c,d))\leq \inf (a,b,c,d)}$.
On the other hand $\inf(a,b,c,d)\leq\inf (a,b) \wedge \inf(a,b,c,d)\leq \inf (c,d)$ (why?). So $\inf(a,b,c,d)$ is a lowerbound for $\{\inf (a,b), \inf (c,d)\}$ which implies that $\color{blue}{\inf (a,b,c,d)\leq \inf (\inf (a,b),\inf (c,d))}$.
By the antisymmetry of $\leq$ the desired result follows.
