How can I prove that $(a,b) = (c,d) \land (c,d)=(e,f) \implies (a,b)=(e,f)$ is true

Question

I am trying to prove this relation, but I just cant. I know it is true, but I can not prove it, because I dont know how. Can someone give me some pointers.

$(a,b) = (c,d) \land (c,d)=(e,f) \implies (a,b)=(e,f)$

Thanks???

EDIT:

This is the part of bigger assingment that I am doing. You can find more here Proving that $(a_1, b_1)\sim(a_2, b_2)\Leftrightarrow\ a_1=a_2\land b_1=b_2$

Bassicly I need to prove that the given relation is transitive, and that is what I am trying to do here in this expression, but I just dont know how to do it.

Answer 1

I suppose you actually want to prove the transitivity for $\sim$ instead of the actual referential equality $=$. So you have to show $$(a,b) \sim (c,d) \land (c,d)\sim(e,f) \implies (a,b)\sim(e,f)$$ assuming that $$(\star)\quad(a_1, b_1)\sim(a_2, b_2)\Leftrightarrow\ a_1=a_2\land b_1=b_2$$ (got that from your other question).

Well, $(a,b) \sim (c,d)$ yields $a=c\wedge b=d$, and $(c,d)\sim(e,f)$ implies $c=e\wedge d=f$. Using the transitivity of $=$ which you may safely assume since $=$ is an equivalence relation, you get $a=c \wedge c=e \implies a=e$ and $b=d \wedge d=f \implies b=f$. Using $(\star)$ in the "$\Leftarrow$" direction, you therefore have the desired $(a,b)\sim(e,f)$.

Answer 2

Hint, assuming

$$(a,b)=(c,d) \iff a=c\wedge b=d\;\;\;\;\;\;(*)$$

then apply $(*)$ once in the $\rightarrow$ direction to break up your pairs on the left hand side, then use the fact that $=$ is transitive:

$$p=q \wedge q=r \implies p=r$$

and finally apply $(*)$ a second time in the $\leftarrow$ direction to form your pairs on the right hand side.

Answer 3

The relation is defined by:

$$\left(a_{1},b_{1}\right)\sim\left(a_{2},b_{2}\right)\iff a_{1}=a_{2}\wedge b_{1}=b_{2}$$

So: $$\left(a,b\right)\sim\left(c,d\right)\wedge\left(c,d\right)\sim\left(e,f\right)\iff\left[a=c\wedge b=d\right]\wedge\left[c=e\wedge d=f\right]$$

So $\left(a,b\right)\sim\left(c,d\right)\wedge\left(c,d\right)\sim\left(e,f\right)$ implies that $a=e\wedge b=f$ or equivalently $\left(a,b\right)\sim\left(e,f\right)$.

Proved is now that $\sim$ is transitive.