document.write( "Question 1171801: I am trying to use indirect proof to derive the conclusion but I am confused on the steps to get to ~A. \r
\n" );
document.write( "\n" );
document.write( "1. (A & B) ⊃C
\n" );
document.write( "2. B & ~C/~A \r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #796713 by Edwin McCravy(20054)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "Start out assuming the negation of the conclusion and reach a contradiction.\r\n" ); document.write( "The conclusion is ~A, so its negation is ~~A\r\n" ); document.write( "\r\n" ); document.write( "A contradiction is always in the form p & ~p. It's C & ~C. We reach it in step\r\n" ); document.write( "10.\r\n" ); document.write( "\r\n" ); document.write( "1. (A & B) ⊃C\r\n" ); document.write( "2. B & ~C/~A\r\n" ); document.write( "\r\n" ); document.write( " | 3. ~~A Assumption for Indirect Proof\r\n" ); document.write( " | 4. A 3, Double Negation\r\n" ); document.write( " | 5. B 2, Simplification\r\n" ); document.write( " | 6. A & B 4,5, Conjunction \r\n" ); document.write( " | 7. C 1,6, Modus Ponens\r\n" ); document.write( " | 8. ~C & B 2, Commutation\r\n" ); document.write( " | 9. ~C 8, Simplification\r\n" ); document.write( " |10. C & ~C 7,9, Conjunction \r\n" ); document.write( " \r\n" ); document.write( "11. ~A Lines 3-10 for Indirect Proof\r\n" ); document.write( "\r\n" ); document.write( "Edwin\n" ); document.write( " |