# identity question

• October 17th 2008, 06:36 AM
inthedl
identity question
Hey guys, so I have the following question...

Suppose we have the operator \$. Let s be a left identity for \$ (s\$x = x for all x) and let r be a right identity for \$ (x\$r = x for all x). Formally prove that s = r.

Here is my approach so far...

Code:

```a \$ s = a a \$ r = a   s \$ r = s \$ r = s       = s \$ r = r So, s = r```
I don't think that this is right. And if it is right, it does not show why. Does anybody have any input as to whether or not what I have is good or how you would go about it? It's important that I show why I do what I do...

Thanks!
• October 17th 2008, 08:45 AM
Opalg
Quote:

Originally Posted by inthedl
if it is right, it does not show why.

That argument is basically right, but it does not clearly show why. You could re-write it like this.

(1) s \$ x = x for all x (definition of s being a left identity).

(2) x \$ r = x for all x (definition of r being a right identity).

Put x = r in (1): s \$ r = r.

Put x = s in (2): s \$ r = s.

Therefore r = s \$ r = s.

That argument would satisfy any practising mathematician. BUT if this question came in a course on formal logic, you would presumably be expected to justify each of the above steps in terms of the axioms of first order predicate calculus (or whatever).
• October 17th 2008, 10:18 AM
inthedl
Thanks very much for your help. While we do have a list of theorems and axioms that we have been using, I believe that what you listed should be enough. I was not able to identify what axioms would explain my logic, but maybe I'll take another look to be safe.
• October 19th 2008, 06:54 AM
ddt
Quote:

Originally Posted by Opalg
That argument would satisfy any practising mathematician. BUT if this question came in a course on formal logic, you would presumably be expected to justify each of the above steps in terms of the axioms of first order predicate calculus (or whatever).

The argument only requires that = is an equivalence relation. Normally, that property is spelled out in a formal logic, isn't it?
• October 21st 2008, 05:44 PM
poutsos.B
Quote:

Originally Posted by inthedl
Thanks very much for your help. While we do have a list of theorems and axioms that we have been using, I believe that what you listed should be enough. I was not able to identify what axioms would explain my logic, but maybe I'll take another look to be safe.

Here is a formal proof:

$\forall x$( s\$x=x) for left identity.......................................... .......................................1

$\forall x$( x\$r=x) for right identity.......................................... .........................................2

s\$r =r: from (1) by using a theorem in predicate calculus called Universal Elimination where we put x=r............................................... ...........3

s\$r=s : ......from (2) and using again Univ. Elim. where we put x=s.........4

s=r :.........by substituting (4) into (3) .Again substitution is a theorem in predicate calculus.......................................... .................................................5

Alternatively we could use the following theorem of equality in predicate calculus to obtain s=r :

$\forall A\forall B\forall C$( A=B & B=C-------> A= C)................................................ .................................................. 6

Now from (6) and using Univ.Elim. where we put A=r we get:
$\forall B\forall C$(r=B & B=C--------> r=C).............................................. .................................................. 7

$\forall C$( r= s\$r & s\$r=C----------> r=C):....from (7) and using again Univ. Elim. where we put B=s\$r............................................. ..............................................8

r= s\$r & s\$r =s ---------> r=s:.....from (8) and using Univ.Elm.where we put C=s............................................... ................................................9

r=s\$r & s\$r=s.........from (3) and (4) and using a law in propositional calculus called Addition Introduction ( p,q =====> p&q , in our case p=(r=s\$r) , q=( s\$r=s) )................................................. ..................10

r=s........from (9) and (10) and using a law in propositional logic called M.Ponens [( p------>q & p)=====>p] .In our case p=( r=s\$r & s\$r=s) ,q=(r=s).

Universal Elimination is alaw where basically says that if: A property holds for the whole set then it will hold for one of its members