Independence of Axioms in Hilbert-Ackermann System The Following are the Axioms used in the Hilbert-Ackerman System:


*

*(X v X) → X

*X → (X v X)

*(X v Y) → (Y v X) 

*(X → Y) → ((Z v X)→(Z v Y))


I need to demonstrate that all of them are independent from each other but I don't know how. 
Edit:
From what I can gather, I need to show the system works with the axioms and the negation of those axioms. So I need to build a system that shows how consistent the system is. Any ideas of how to build said system?
Thanks for any help!
 A: One way to show that, say, 2 is logically independent from 1,3, and 4, is to provide a semantics such that 1,3, and 4 have a semantical property that 2 does not have. Since all statements are theorems in classical propositional logic, this semantics will have to be something that is completely different from truth-functional semantics.
For example, suppose that the variables in the expressions can take on one of the values $A$, $B$, or $C$, and suppose that the operators $\lor$ and $\rightarrow$ map a pair of these values to another one of those values ( so again, this has nothing to do with good old truth-functional logic anymore, where operators map truth-values to truth-values ... here, they simply map $A$'s, $B$'s, and $C$'s to each other). 
Furthermore, suppose that, as it so happens, statements 1,3, and 4 always end up mapping to the value $A$ (we could therefore call them '$A$-necessities'), but statement 2 is not an $A$-necessity. 
Finally, suppose that with the inference rules that you have (I suppose your system that is only Modus Ponens?), you can only infer $A$-necessities from other $A$-necessities. Then it should be clear that, starting with any instances of axioms 1,3, and 4, you can never reach axiom 4, applying the inference rules that you have, and thus 4 is idependent from the others.
So, this is what you can try to do: try to construct a table as you would a normal truth-table, but now with all 9 possible pairs of $A$, $B$, and $C$ for both the $\lor$ and the $\rightarrow$, such that all of the above happens.... this could take a while ... maybe write a computer program to go through all possibilities ... and even then, it may not work; but if it doesn't work, maybe try a 4-value system ... Good luck!
