1. (G&H)>(J<>L)

2.  G<>H

3.  (H&~L)v(H&K)

Conclusion: J>K 

----------------------------
Using a conditional proof, we assume the truth of J, 
so we put a "t" over all J's
--------------------

           t
1.  (G&H)>(J<>L)

2.  G<>H

3.  (H&~L)v(H&K)

            t 
Conclusion: J>K 

---------------------

Since J<>L appears in 1, and J is t, then we must put a t above all Ls

---------------------

           t  t
1.  (G&H)>(J<>L)

2.  G<>H

        t
3.  (H&~L)v(H&K)

            t 
Conclusion: J>K 

---------------------
Since L has a t over it, in the ~L of 3, we put an f over the ~ 


           t  t
1.  (G&H)>(J<>L)

2.  G<>H

       ft
3.  (H&~L)v(H&K)

            t 
Conclusion: J>K 

---------------------
Since there is an f over the ~L in 3, there is also an f
over the & of (H&~L).


           t  t
1.  (G&H)>(J<>L)

2.  G<>H

      fft    
3.  (H&~L)v(H&K)

            t 
Conclusion: J>K 

---------------------

Since the left side of 3. (H&~L) has an f over it, the right
side must have a t over it, so that 3 will be a true premise,
so we put a t over the & of H&K 

           t  t
1.  (G&H)>(J<>L)

2.  G<>H

      fft t  t 
3.  (H&~L)v(H&K)

            t 
Conclusion: J>K 

---------------------

Since the & of H&K has a t over it, both H, K must have t's over them.
So we put t's over all K's
 
           t  t
1.  (G&H)>(J<>L)

2.  G<>H

      fft t ttt 
3.  (H&~L)v(H&K)

            t t 
Conclusion: J>K 

---------------------

Therefore the conclusion is valid, because the assumption of the 
truth of J leads to the truth of K.

Edwin