document.write( "Question 162437This question is from textbook
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Algebra.Com's Answer #119709 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "(P & Q) <=> ~(P -> ~Q)\r\n" );
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document.write( "Using the rules:\r\n" );
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document.write( "Under P put TTFF,\r\n" );
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document.write( "Under Q put TFTF,\r\n" );
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document.write( "The rule for \"~\" is \"~T is F and ~F is T\",\r\n" );
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document.write( "The rule for \"&\" is \"only T&T is T, all others F\",\r\n" );
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document.write( "The rule for \"V\" is \"only FVF is F, all others T\",\r\n" );
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document.write( "The rule for \"->\" is \"only T->F is F, all other T\",\r\n" );
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document.write( "The rule for \"<->\" is \"only T<->T and F<->F are T, all others F,\r\n" );
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document.write( "make this truth table:\r\n" );
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document.write( "| P | Q | ~Q | P & Q | P -> ~Q | ~(P -> ~Q) | (P&Q) <-> ~(P -> ~Q} |\r\n" );
document.write( "| T | T |  F |   T   |    F    |     T      |        T             | \r\n" );
document.write( "| T | F |  T |   F   |    T    |     F      |        T             |\r\n" );
document.write( "| F | T |  F |   F   |    T    |     F      |        T             |\r\n" );
document.write( "| F | F |  T |   F   |    T    |     F      |        T             |\r\n" );
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document.write( "The proposition is proved because there are only T's in the last \r\n" );
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document.write( "Therefore we can replace the biconditional symbol <->, by the\r\n" );
document.write( "stronger equivalence symbol <=> and write\r\n" );
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document.write( "(P&Q) <=> ~(P -> ~Q}\r\n" );
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document.write( "Edwin
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