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For an identity relation, are the following mathematical notations correct?

∀a,b ∈ A : aRa ∧ ∄ aRb : a ≠ b

R = { (a,a) : a ∈ A }

What i want to say in first notation is for every a and b in A, a is related to itself, and there does not exist a relation between a and b, where a is not equal to b.

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  • $\begingroup$ Hi, welcome. Can you include in your question the statements that you are trying to express in notation? The second looks syntactically correct, I'm just not sure what you mean. The first I am having trouble parsing. $\endgroup$ – Matthew Leingang Sep 21 '18 at 13:33
  • $\begingroup$ Oh, I think I missed the title of the question. That helps. $\endgroup$ – Matthew Leingang Sep 21 '18 at 13:35
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The second one is definitely correct. You can define a relation by its pairs. $$ R = \left\{(a,a) : a \in A \right\} $$

Reading the first one, it looks like you're saying that $a$ is related to $a$ for every $a$, and there is no $b$ not equal to $a$ that is related to $a$. That might look more like this: $$ \forall a \in A,\left( a \mathrel{R} a\right) \wedge \neg \left(\exists b \in A,\ (a \neq b) \wedge (a \mathrel{R} b)\right) $$ That second part can be simplified logically, though. \begin{align*} \neg \left(\exists b \in A,\ (a \neq b) \wedge (a \mathrel{R} b)\right) &\equiv \forall b \in A, (a=b) \vee (a \not\mathrel{R} b) \\ &\equiv \forall b \in A, (a \mathrel{R} b) \implies (a = b) \end{align*} That last part is because $p \implies q$ is equivalent to $(\neg p) \vee q$.

The first part can be combined with it: \begin{align*} \forall a \in A,\ a \mathrel{R} a \equiv \forall a \in A,\ \forall b \in A,\ \left(a = b \implies a \mathrel{R} b\right) \end{align*} So the two together say: \begin{align*} &\forall a \in A,\ \forall b \in A,\ \left((a = b) \implies (a \mathrel{R} b)\right) \wedge \left((a \mathrel{R} b) \implies (a = b)\right) \\&\equiv\forall a \in A,\ \forall b \in A,\ \left((a = b) \iff (a \mathrel{R} b)\right) \end{align*}

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  • $\begingroup$ Please look at it again, i made some edits. $\endgroup$ – Tayyab Mazhar Sep 21 '18 at 14:21
  • $\begingroup$ so, p ⟹ q is false when p is true and q is false and, p ⟺ q is false when p is true and q is false, or p is false and q is true. So considering your last notation, R is identity if ((a=b)⟺(aRb)) is true for every pair of AxA. Am i correct? $\endgroup$ – Tayyab Mazhar Sep 21 '18 at 15:12

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