Well I have been playing with this problem for about a week. I wonder if you have the time to show me how to resolve, my problems. There are three conditions where 0—>dim(L# conditions can exist. That is when L# have no value

1) L₁

2) L₅

3) both L₁ and L₅

For each condition of the values of L₁ and L₅ there is only one output, which is L₃ After the first If statement there is a prgmCALL. I have just used that OUTPUT statement to simulate that, since when the L# contain values there is no problem and the real action is in the else bucket. I just change the conditions for the L# to test my results.

{5,6,7,8}→L₁

{1,2,3,4}→L₅

ClrHome

If 0≠dim(L₁ and 0≠dim(L₅

Then

Output(1,1,"FIRST L3 "L₃

Else

If 0≠dim(L₁ and 0=dim(L₅:L₁→L₃

Output(2,1,"SECOND L3 ",L₃

If 0≠dim(L₅ and 0=dim(L₁:L₅→L₃

Output(3,1,"THIRD L3 ",L₃

End

I await your reply with great anticipation.:-)

john