We let $(A,\leq)$ be an ordered set. We define a relation $\preceq$ on AxA by: $$(a_1,a_2)\preceq(b_1,b_2) <=>[(a_1\neq b_1) ∧(a_1\leq b_1)]∨ [(a_1=b_1)∧(a_2\leq b_2)]$$ I think we have that $(a_1\neq b_1) ∧(a_1\leq b_1)->a_1<b_1$. So we can reduce it to: $$(a_1,a_2)\preceq(b_1,b_2) <=>[a_1<b_1]∨ [(a_1=b_1)∧(a_2\leq b_2)]$$ Then I have to show that $\preceq$ is an order-relation. I have shown that we have reflextivity and transitivity. But I got some problem to show that we got antisymmetry.
If I check for the second condition it works while: $$((a_1=b_1)\wedge(a_2\leq b_2))\wedge((b_1=a_1)\wedge(b_2\leq a_2)) =>(a_1,a_2)=(b_1,b_2)$$ But if I check for the first condition I got: $$(a_1<b_1)\wedge(b_1<a_1) => emty?$$ Can someone help me proving the antisymmetry?
For antisymmetrie, consider the premise
$[(a_1=b_1)\wedge (a_2≤b_2)]\wedge [(b_1=a_1)\wedge (b_2≤a_2)]$.
Since $a_2\leq b_2 \wedge b_2\leq a_2$ is equivalent to $a_2=b_2$, it follows that $(a_1,a_2) = (b_1,b_2)$.