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Is the relation $R$ on $A=$ the set of all word of English, defined by

$R=\{(x,y)\in A\times A: $ the first letter of the word $y$ occurs at least as late in the alphabet as the first letter of the word $x \}$ a partial order?

I said no because $(\text{ball},\text{bass}) \in R$ and $(\text{bass},\text{ball})\in R$, but $\text{bass} \neq \text{ball}$. Hence, not antisymmetric, but I read elsewhere that it is antisymmetric.

M47145
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Jing Weng
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  • Your question is unclear as stated. But this just creates equivalence classes of words by their first letter. Intuitively, that sounds like a partial order to me. apple $=$ arrogant as far as the first letter is concerned. – jdods May 16 '16 at 22:46

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Your proof does indeed show that $R$ is not a non-strict partial order. Perhaps what you read elsewhere claimed that $R' = \{(x,y) ~|~ \text{the first letter of }x\text{ occurs later than the first letter of }y\}$ is a strict partial order, which is of course true.

2xThink
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