prove that if $a=b$ then $a+c=b+c$ where $a,b,c\in \mathbb R$ I was trying to prove if $l=m$ and $m=n$ then $l=n$ but when doing this I had to add $-m$ to both sides of both equations.i think it is not appropriate to proceed without proving "if $a=b$ then $a+c=b+c$". can anyone help? I tried many ways but every time I failed.
note: in this context when prooving $a=b \implies a+c=b+c$ one can not use 
$l=m$ and $m=n  \implies l=n$
 A: Here is the thing, if you are starting from $\mathbb R$ defined by axioms of complete ordered field, or even just axioms of field, there are two operations that are defined on $\mathbb R$, addition and multiplication. By definition of binary operation, $+\colon \mathbb R\times\mathbb R\to \mathbb R$ is a function. It follows that for every $c\in\mathbb R$, $f_c\colon \mathbb R\to\mathbb R$, $f_c(x) = x + c$ is a function as well. Now remember that one of the defining properties of a function is the following $$a = b\implies f(a) = f(b)$$
and apply it to $f = f_c$ to get what you want. Thus, it is already encoded in the axioms just by saying that $+$ is binary operation. The same goes for multiplication.
A: Equality has the property one can replace one half of the equality with the other half in any term where the order of operations is clear.  I'll call this the replacement property of equality.
Suppose that a = b.  Then, by the replacement property of equality, specifically by replacing the 'b' with 'a', it follows that a = a.  Now 'a' was an arbitrary real number.  Thus, for any real number x, 
x = x.
Since '+' is an operation, (a + c) is some real number.  By the lemma x = x, it thus follows that 
(a + c) = (a + c).  
Using the replacement property of equality and the hypothesis yields the result of the question.
The replacement property of equality also suffices to prove that if "l = m, and m = n, then l = n."
A: There isn’t much to prove. If $a=b$, then $a+c=b+c$ by direct substitution. The converse is more interesting to prove. 
That is, if $a+c=b+c$, then $a=b$.
Consider $a=a+0=a+(c+(-c))=(a+c)+(-c)=(b+c)+(-c)=b+(c+(-c))=b+0=b$. Thus $a=b$.
A: So, proving the fact that $(l = m \land m = n) \implies l = n $  does not require knowing anything about $+$, it is the transitive property of equality, which is an axiom.

Let's prove the statement in the title anyway.
So, we have three variables, all universally qualified.
$$ (a = b) \implies (a + c = b + c) \;\forall a \forall b \forall c $$
$$ (a \ne b) \lor (a + c = b + c) \;\forall a \forall b \forall c $$
Negate the goal.
$$ (a = b) \land (a + c \ne b + c) \;\exists a \exists b \exists c \tag{NG}$$
We can add three constants to our context. Note that by doing so we are extending the theory of the real numbers to include three new constants $a, b, c$ that collectively satisfy the property we want.
$$ a = b \;\; \text{and} \;\; a+c \ne b + c\tag{101} $$
$$ a = b \tag{102} $$
The step $102 \to 103$ only requires noticing that $t \mapsto t + c$ is a function. Every function preserves equalities. A non-injective function may introduce new equalities, but no function can remove them.
$$ a + c = b + c \tag{103} $$
Pick the left expression of (101).
$$ a + c \ne b + c \tag{104} $$
(104) is the negation (103) syntactically. 
$$ \bot $$
