Symbolic Logic Proof? I was given the following argument: 
If there is a, there is b and if there b there is c. Thus there is a but no c.
Which I put into symbolic form:
(a $\rightarrow$ b) $\land$ (b $\rightarrow$ c)
$\therefore$ a $\land$ $\lnot$ c
I am supposed to either establish the validity of this argument with a proof or show that it is invalid by providing a counterexample. I have looked over the rules of inference and laws of logic but I did not see any rule or law that I can apply. I believe that it is invalid but am really stuck on how to provide the counterexample please help.
 A: If $a,b,c$ all true, then premises true and conclusion false.
A: 
I have looked over the rules of inference and laws of logic but I did not see any rule or law that I can apply. 

Well, that is suggestive.

I believe that it is invalid but am really stuck on how to provide the counterexample please help.

The counter example can be found when the conclusion is false; which is to say, when $\neg a$ or $c$. 
So, if the premise can be satisfied for either of these, then the argument is invalid.
So can $(a\to b)\land(b\to c)$ be true when $a$ is false?   What about when $c$ is true?
A: The problem the OP has is how to find a counterexample. This answer will provide three ways.


*

*Create a statement with the premises conjoined (and-ed) together and have that imply the conclusion. Then construct a truth table based on that statement as follows:



Note the column under the implies sign (→). There are two rows with "F" in that column. Each of these rows will allow one to provide a counterexample. To note the counterexample specify the valuations for the variables in the row that make that column false.  
For this truth table there are two counterexamples: 

$A$ is false, $B$ is true and $C$ is true (fifth row)
$A$ is false, $B$ is false and $C$ is true (seventh row) 



*Use a tree proof generator. Look for branches that do not close. From those branches one can construct counterexample. This tool generates a counterexample (countermodel) if the tree does not close.



Note that this countermodel corresponds to the seventh row in the above truth table.


*For a counterexample to exist, the conclusion must be false while the premises are true. Knowing this we can try to manually construct a counterexample using the following steps:


*

*First, find the valuations of $A$ and $C$ that will make the conclusion $A\lor\neg C$ false? The only way that can happen is if $A$ is false and $C$ is true.

*Second, setting $A$ as false and $C$ as true is there a value of $B$ that will make all of the premises true? As it turns out any value of $B$ would make the premises true. 

*Third, since we found values we can present a counterexample. 




Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html
Tree Proof Generator. https://www.umsu.de/logik/trees/
