SOLUTION: Conditional proof M->(K->L) (L\/N) -> J Therefore, M-> (K->J) That's supposed to be a downward arrow between L and N in the second line

1. M-->(K-->L)      Premise

2. (LvN)--> J       Premise

// show M-->(K-->J)

3.:: M              Conditional Proof (CP) assumption #1

4.:: K-->L          3,1  Modus Ponens (MP)

5.:: K              CP assumption #2

6.:: L              5,4  MP

7.:: LvN            6  addition (ADD)  [ The free lunch of logic ]

8.:: J              7,2  MP

9.:: K-->J          5-8  CP

10.:: M-->(K-->J)   3-9  CP

11. M-->(K-->J)     3-10 CP  



// Notes

Line 3  says assume M is true

Line 4  shows if M is true then it follows  K-->L  by Premise #1

Line 5  says assume K is true

Line 9  shows the if K is true then J is true (K-->J)

Line 10 just puts together "if M true then it follows K-->J"  so M-->(K-->J)

Line 11 discharges CP assumptions and brings the conclusion into the main

         argument.