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I would like to express that sentence as $$\forall x,y\in K(xRy \lor yRx)$$ but how can I say that $x\neq y$? Like this? $$\forall x,y \in K((xRy \lor yRx) ∧ x\neq y)$$

amWhy
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  • On the use of $\land$ vs. $\to$ in quantification, see also here: https://math.stackexchange.com/questions/2260428/why-can-we-use-implication-with-the-universal-quantifier-but-not-with-the-existe – Natalie Clarius Feb 29 '20 at 17:27
  • If one of the answers below are satisfying for you, please accept it. – Mauro ALLEGRANZA Mar 05 '20 at 12:33

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No, the last statement is false because nothing in the quantifier prohibits $x=y$. You can say $$\forall x,y∈K~(x \neq y \to (xRy ∨ yRx))$$ which says nothing about whether $xRx$ or not. I think I have also seen $$\forall \underset{\large x\neq y}{x,y∈K}~(xRy ∨ yRx)$$

Graham Kemp
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Ross Millikan
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As a general rule of thumb, a statement of the form

For all $x \in X$ which are $P$, $Q$ holds

translates as

$\forall x \in X (P \to Q)$

You can paraphrase your sentence as

For all $x, y \in K$ which are distinct, $xRy$ or $yRx$ holds

so the formalization becomes

$\forall x, y \in K (x \neq y \to (xRy \lor yRx))$


Your attempt is not an adequate formalization because it entails that all $x, y \in K$ must be distinct from each other (since $xRy \lor yRx$ and $x \neq y$ holds for all of them), but since the universal quantififier ranges over any possible combiniation of $x, y$, your formula must also be true for $x = y$, which is contradictory.
With the implication instead, the formula just says that if they are distinct, then $xRy$ or $yRx$ should hold, but if not, we don't care.