
Ordered Triple Proof
This is how I've defined an ordered triple: ((a,b),c) is the set { {{a,1},{b,2}} , {c,3} }. Based on the fact that I've already proved equality for ordered pairs (a,b) = (c,d) iff a=c, b=d, I'm trying to prove the same for the triples. This is what I've got:
((a,b),c) = ((d,e),f) if
{ {{a,1},{b,2}} , {c,3} } = { {{d,1},{e,2}} , {f,3} }
These sets are equal if:
i) {{a,1},{b,2}} = {{d,1},{e,2}}, which is true iff a=d, b=e
and {c,3} = {f,3}, which is true iff c=f.
Thus, in this case, a=d, b=e, c=f.
ii) {{a,1},{b,2}} = {f,3} and {c,3} = {{d,1},{e,2}}
What do I do about this second case? Can I ignore it because it is impossible for them to be equal? I can't figure out how to deal with it. Is there a problem in my definition for an ordered triple? I know my definition for an ordered pair is different from the usual {{a},{a,b}} but that is what my book is using. Any help would be appreciated!
Thanks,
Ultros

Say $\displaystyle (a,b,c) = (d,e,f)$.
By definition $\displaystyle ((a,b),c) = ((d,e),f)$.
But you already proved this for pairs.
Thus $\displaystyle (a,b)=(d,e)$ and $\displaystyle c=f$.
Using pairs again $\displaystyle a=b$ and $\displaystyle b=e$.