False(ified) Axioms Don't really know how to title this one. I'm working on a real analysis question that says: 

In one sentence write down the reason why $(1=2)\land (2=3)\to (1=3)$ (and similar substitutions) don't lead to false statements when we use the transitivity axiom in other proofs?

The question references a previous question wherein it defines the transitivity axiom to be $\forall x\forall y\forall z (x=y)\land(y=z)\to(x=z)$, but then states that if we choose $x=1,y=2,z=3$ the theorem gives $1=3$. 
The question also invites us to consider how modus ponens is used in proofs.
The idea I have in my mind is that for the theorem to be used first the values we substitute must hold true for the theorem i.e. because the theorem states $x=y$, but the values we choose have $x\ne y$ then the theorem cannot be used. But I'm not convinced this is concise enough or that it holds for other proofs when using this and other axioms. With regards to modus ponens, my understanding was that if we assume $A$ to be true and $ A\to B$ then we can infer $B$ is true, but I'm not 100% sure how this can be applied when 3 values are chosen somewhat at random to be used.
Any help would be greatly appreciated.  
 A: This is an example of the need to be careful in reading logic statements. In this statement, what it says is that
"if 1=2 and 2=3, then 1=3". Does 1=2? No. Therefore, the statement isn't false. So long as $1\neq 2$ or $2\neq 3$, this statement will be true by default.
Kind of like saying "if pigs fly, then I'll be a monkey's uncle". That statement is true, not because I'm a monkey's uncle, but because pigs don't fly.
A: As Glen points out, the whole statement:
$$\text{$\bf If$}\;\;1=2\;\;\text{and}\;\; 2= 3,\;\;\text{$\bf then$}\;\; 1 = 3\tag{1}$$ 
is true, for the reasons Glen gives. We have $F\implies F$, which evaluates as true.
However, it does not follow that $(2 = 3)\;\;$ is true. I.e., $(1)$ does not give us $1 = 3$. It gives us only:
$$\text{$\bf If$}\;\;1=2\;\;\text{and}\;\; 2= 3,\;\;\text{$\bf then$}\;\; 1 = 3\tag{1}$$ 
Why does it not give us that in fact, $1 = 3$? Here is where modus ponens comes to the rescue (or, rather, the inability to validly apply modus ponens saves the day). 
To conclude therefore $1 = 3$, we would need the truth of $1 = 2 \land 2 = 3$, together with $(1)$, and by applying modus ponens, get $1 = 3$. 
But we don't have the truth of $1 = 2 \land 2 = 3$, so we cannot apply modus ponens to conclude that in fact $1 = 3$.
So we are left with nothing more than that the following proposition, as a whole, is true:
$$\text{$\bf If$}\;\;1=2\;\;\text{and}\;\; 2= 3,\;\;\text{$\bf then$}\;\; 1 = 3$$
and from this alone, we cannot conclude that $\;1 = 3$ is true
A: Since $1\neq 2$ and $2\neq 3$, you would never be able to apply $(1=2)\wedge (2=3)\rightarrow 1=3$ so no harm done by having it as an axiom or theorem.
