Conceptual question about how one procedurally establishes equality between two "novel" objects.

Question

After reading the following post (Why does one have to check if axioms are true?), I wanted a reality check to make sure that I am thinking about equivalence relations correctly.

In Tao's Analysis I, the following statements are made:

Two ordered pairs $(x,y)$ and $(x',y')$ are considered equal if and only if both their components match, i.e.$(x,y) = (x',y') \iff x=x' \text{ and } y=y'$. This obeys the usual axioms of equality (Exercise 3.5.3).

Exercise 3.5.3 reads as follows:

"Show that the definitions of equality for order pair [...] obey the reflexivity, symmetry, and transtivity axioms".

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I am looking for confirmation that the following thinking is the correct sequential process one goes through when determining equality between two objects.

Imagine that I just discovered/invented the concept of an ordered pair and don't know anything about them. I subsequently state the following:

$(x,y) R_{op} (x',y') := (x=x' \land y=y')$, where the parentheses on the right are strictly being used to "separate" the symbology from the "defined as" symbol, :=.

i.e. I am first defining what the relation $R_{op}$ means. (where "op" is short for ordered pair).

It dawns on me that this relationship that I just defined could, possibly, satisfy the 3 axioms of equality. I go through and confirm these three statements:

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Reflexivity: $(x,x) R_{op} (x,x)$

1. $(x=x \land y=y)$

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Symmetry: if $(x,y) R_{op} (x',y')$, then $ (x',y') R_{op} (x,y)$

2. if $(x=x' \land y=y')$, then $(x'=x \land y'=y)$

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Transitivity: if $(a,b)=(c,d) \text{ and } (c,d)=(e,f)$ then $(a,b) = (e,f)$

3. if $(a=c \land b=d) \land (c=e \land d=f)$, then $(a=e \land b=f)$

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I then say, "Oh, man. This relationship satisfies all three axioms of equality"

Consequently, I can "promote" $R_{op}$ to an equivalence relation, which allows me to use the symbol "$=$" instead of "$R_{op}$". Now, and only now, can I write what Tao writes, which is:

$(x,y) = (x',y') \iff x=x' \text{ and } y=y'$

And, really, I think the more pedantic way of formulating this is:

$(x,y) = (x',y') \iff (x,y) R_{op} (x',y')$, where $(x,y) R_{op} (x',y') := (x=x' \land y=y')$

Is this procedurally correct? Thank you!

Answer 1

Reflectivity
(x,y) = (x,y) iff x = x and y = y

Symmetry (x,y) = (a,b) iff x = y and a = b
iff a = b and x = y iff (a,b) = (x,y)

Transitivity. Assume (a,b) = (u,v) and (u,v) = (x,y).
Thus a = u and b = v and u = x and v = y.
Hence a = x and b = y; (a,b) = (x,y).