 # I am stuck on 3 proofs for my discrete math class. Any help would be greatly appreciated. 1. Prove that if R is a partial order, then R^{-1} is a partial order. 2.Prove that if R_1 and R_2 are equivalence relations on the set A, then R_1 cap R_2 is an equivalence relation. Maliyah Robles 2022-07-16 Answered
I am stuck on 3 proofs for my discrete math class. Any help would be greatly appreciated.
Prove that if R is a partial order, then ${R}^{-1}$ is a partial order
Prove that if ${R}_{1}$ and ${R}_{2}$ are equivalence relations on the set A, then ${R}_{1}\cap {R}_{2}$ is an equivalence relation.
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Step 1
Here is part of (1). The rest of (1) is similar, so is (2). Problem (3) is really a completely different topic, I suggest you delete it and ask a separate question.
Let R be a partial order: therefore R is reflexive, transitive and antisymmetric. We prove that ${R}^{-1}$ is transitive.
Step 2
So, suppose that $x\phantom{\rule{thinmathspace}{0ex}}{R}^{-1}\phantom{\rule{thinmathspace}{0ex}}y$ and $y\phantom{\rule{thinmathspace}{0ex}}{R}^{-1}\phantom{\rule{thinmathspace}{0ex}}z$. By definition of inverse this means that yRx and zRy. Since R is transitive we have zRx, and using the definition of inverse again, $x\phantom{\rule{thinmathspace}{0ex}}{R}^{-1}\phantom{\rule{thinmathspace}{0ex}}z$. We have proved that if $x\phantom{\rule{thinmathspace}{0ex}}{R}^{-1}\phantom{\rule{thinmathspace}{0ex}}y$ and $y\phantom{\rule{thinmathspace}{0ex}}{R}^{-1}\phantom{\rule{thinmathspace}{0ex}}z$ then $x\phantom{\rule{thinmathspace}{0ex}}{R}^{-1}\phantom{\rule{thinmathspace}{0ex}}z$; by definition, ${R}^{-1}$ is transitive.
Observe that this proof really uses pretty much nothing except various definitions. So I hope this underlines the importance of knowing the definitions properly.

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