The symbol "-A" is not mentioned in your axioms so you can't prove "A+ (-A)= 0"! How are you defining "-A"? Do you not mean that you want to prove that "for all A there exist B such that B+ A= 0"? In that case, yes, it follows from (4) and (1).
Given the following axiomatic system :
For all A,B,C:
1) A+B= B+A .................................................. ......AB= BA
2) A+(B+C)= (A+B)+C........................................... ....(AB)C= A(BC)
3) A+0 =A................................................ .................1A=A
4) For all A ,there exists B : A+B=0.....................................For all A =/=0 ,there exists B: AB=1
5) A(B+C)= AB+AC
6) 1=/=0
We want to prove:
For all A : A+(-A) = 0
Proof:
From axiom (4) if we put A = -B we have : (-B)+B =0 => B+(-B) =0 ,by axiom (1)
Hence by changing the variables we have :
For all A : A+(-A) = 0
I s that proof correct??
The symbol "-A" is not mentioned in your axioms so you can't prove "A+ (-A)= 0"! How are you defining "-A"? Do you not mean that you want to prove that "for all A there exist B such that B+ A= 0"? In that case, yes, it follows from (4) and (1).
Where do you base that.
On the other hand i can claim the opposite supported by the following axiom of the predicate logic:
, where P is a formula, t is a term and u is a variable
And in our case we have: , where -B is a term and A a variable and .
Hence (-B) + B = B+(-B)=0
You cannot prove anything about "-A" because you have not defined "-A". I was not clear whether "A", "B", and "C" were integers and "+" addition or if "A", "B", and "C" were "statements" and "+" is "and". In either case, you have to define "-A" explicitly.