Math is objective

1. P is a theorem in F. That is, using only the deduction rules of F, we can deduce P from the axioms of F.
   
2. P is not a theorem in F. That is, not such thing is possible.
   

Reading into math has been interesting.  Thanks for the debate topic and chance to talk about it. 

A ) Is formal system objective?

B)  What would you say is the "language" refered to in definition/description provided for formal systems, and would this language be objective? 

C) The Zermelo-Fraenkel axioms are the basis for Zermelo-Fraenkel set theory. In the following (Jech 1997, p. 1), 

 stands for exists,  means for all,  stands for "is an element of,"  for the empty set,  for implies,  for AND,  for OR, and  for "is equivalent to."

A)  Are these symbols and meanings used through out the world? Do you know what existed or was used before these symbols were promoted by Ermelo? 

B) there are and/or have been different formulas & methods used to find the same answer. A common running joke in media is a parent trying to help their kid do math but realizing the process to find the answer is different. 

Ancient societies like methods used by Byzantines had similar mathmatical processes to today's standard. Would a change in formula or method to determine an answer demonstrate subjectivity in math because the  formulas/methods in doing math change? 

In your given presentation for F  and P. You give only two options. These options are 

i p is in f

ii p is not in f

Is this accurate depiction for your presentation? 

2 + 2 = 4 ..  yes? Can only be 4 then. That's objective. 

Here is a good one. 

0 = 0

Lets add some numerical values so zero is zero .

4-4 = 10-10

We can rewrite this as a product of 2

2*2 & 2*5

Lets put it into a solving equation. 

(2-2)(2+2)= 5(2-2). 

Interesting enough. What if we cancel (2-2) from each side?

 

We get

(2+2)=5

(2-2)(2+2)= 5(2-2). 

Interesting enough. What if we cancel (2-2) from each side?

 

We get

(2+2)=5

If we use a modulo system containing restricted numbers similar to above clock example, we can develop equation 2+2= 1. 1 represents 4 in this system.  

Even if we look at equation 4-4 = 5-5

Use distribution property. 

4(1-1)= 5(1-1)

Cancel (1-1) 

To get 

4=5 

You would say this is incorrect? 

Why is this the case? 

1-1 is equal to zero just like 2-2. You cancelation property does not apply to zero. Zero is zero no matter how you write it.

Why? 

This seems to be the accepted method, but is there information into why cancelation property does not apply to zero? 

How do we know if this is not a long line of circular reasoning? 

Unfortunately for you, your use of it is invalid, as you attempted to cancel 2-2 from both sides, which is equal to 0. Now, I could stop right here, but for any who feel that this is just mathematicians putting arbitrary rules in place, here is where the cancelation property comes from:

Suppose ab=ac. Then we subtract ac from both sides: ab-ac=0. Now we apply the distributive property: a(b-c)=ab-ac=0. So we see that a(b-c)=0. When two things multiply to zero, one of them must itself be equal to zero, so either a=0 or b-c=0. At this point we can conclude that if a is not equal to 0, than b-c must equal 0, and so b=c. This, however, requires that a be non-zero, as if a=0, than there is no need for b-c to be equal to zero, and so b is not necessarily equal to c.

Unfortunately for you, your use of it is invalid, as you attempted to cancel 2-2 from both sides, which is equal to 0. Now, I could stop right here, but for any who feel that this is just mathematicians putting arbitrary rules in place, here is where the cancelation property comes from:

Suppose ab=ac. Then we subtract ac from both sides: ab-ac=0. Now we apply the distributive property: a(b-c)=ab-ac=0. So we see that a(b-c)=0. When two things multiply to zero, one of them must itself be equal to zero, so either a=0 or b-c=0. At this point we can conclude that if a is not equal to 0, than b-c must equal 0, and so b=c. This, however, requires that a be non-zero, as if a=0, than there is no need for b-c to be equal to zero, and so b is not necessarily equal to c.