$A,B$ ve $A\cup B$ alttan sınırlı ve $x\in A\cup B$ olsun.
$\left.\begin{array}{rr} A\subseteq A\cup B \Rightarrow \inf(A\cup B)\preceq\inf A \\ \\ B\subseteq A\cup B \Rightarrow \inf(A\cup B)\preceq\inf B \end{array}\right\}\Rightarrow \inf(A\cup B)\preceq \min\{\inf A,\inf B\} \ ...(1)$
$\begin{array}{rcl} x\in A\cup B & \Rightarrow & x\in A \ \vee \ x\in B \\ \\ & \Rightarrow & \inf A\preceq x \ \vee \ \inf B\preceq x \\ \\ & \Rightarrow & \min\{\inf A,\inf B \}\preceq x \\ \\ & \Rightarrow & \min\{\inf A,\inf B \}\in (A\cup B)^a \\ \\ & \Rightarrow & \min\{\inf A,\inf B \}\preceq\inf(A\cup B) \ ...(2) \end{array}$
$(1) \text{ ve } (2) \Rightarrow \inf(A\cup B)=\min\{\inf A,\inf B\}.$
O halde söz konusu önerme doğrudur.