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)\"\" \"About 
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( "
\n" );