document.write( "Question 1120001: 1. T ⊃ B Prove: L • B (Hint this involves using the Addition Rule)
\n" ); document.write( "2. T ⊃ ~X
\n" ); document.write( "3. T
\n" ); document.write( "4. X
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #735645 by math_helper(2461)\"\" \"About 
You can put this solution on YOUR website!
1. T ⊃ B Prove: L • B (Hint this involves using the Addition Rule)
\n" ); document.write( "2. T ⊃ ~X
\n" ); document.write( "3. T
\n" ); document.write( "4. X \r
\n" ); document.write( "\n" ); document.write( "—————————————————————————————————
\n" ); document.write( "There is a logical contradiction in any attempted proof. Look at it. \r
\n" ); document.write( "\n" ); document.write( "

\n" ); document.write( "1. T—>B Premise
\n" ); document.write( "2. T—>~X Premise
\n" ); document.write( "3. T Premise
\n" ); document.write( "4. X Premise
\n" ); document.write( "5. ~X 3,2 Modus Ponens (MP)
\n" ); document.write( "6. X & (~X) 5,4 Contradiction — we must stop here!
\r
\n" ); document.write( "\n" ); document.write( "Since the attempted proof contains a contradiction, one can not make a sound proof of L & B. Note that another way to look at this: it may be possible to \"prove\" L & B but the proof will not be sound. This is similar to proving 2=1 when there is a division by zero somewhere in the chain of reasoning. \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );