document.write( "Question 1177235: Conditional proof
\n" ); document.write( "M->(K->L)
\n" ); document.write( "(L\/N) -> J
\n" ); document.write( "Therefore, M-> (K->J)\r
\n" ); document.write( "\n" ); document.write( "That's supposed to be a downward arrow between L and N in the second line
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Algebra.Com's Answer #805710 by math_helper(2461)\"\" \"About 
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\n" ); document.write( "1. M-->(K-->L) Premise
\n" ); document.write( "2. (LvN)--> J Premise
\n" ); document.write( "// show M-->(K-->J)
\n" ); document.write( "3.:: M Conditional Proof (CP) assumption #1
\n" ); document.write( "4.:: K-->L 3,1 Modus Ponens (MP)
\n" ); document.write( "5.:: K CP assumption #2
\n" ); document.write( "6.:: L 5,4 MP
\n" ); document.write( "7.:: LvN 6 addition (ADD) [ The free lunch of logic ]
\n" ); document.write( "8.:: J 7,2 MP
\n" ); document.write( "9.:: K-->J 5-8 CP
\n" ); document.write( "10.:: M-->(K-->J) 3-9 CP
\n" ); document.write( "11. M-->(K-->J) 3-10 CP \r
\n" ); document.write( "\n" ); document.write( "// Notes
\n" ); document.write( "Line 3 says assume M is true
\n" ); document.write( "Line 4 shows if M is true then it follows K-->L by Premise #1
\n" ); document.write( "Line 5 says assume K is true
\n" ); document.write( "Line 9 shows the if K is true then J is true (K-->J)
\n" ); document.write( "Line 10 just puts together \"if M true then it follows K-->J\" so M-->(K-->J)
\n" ); document.write( "Line 11 discharges CP assumptions and brings the conclusion into the main
\n" ); document.write( " argument.\r
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