Algebra.Com's Answer #107986 by Edwin McCravy(20056)  You can put this solution on YOUR website! Can someone help me construct this truth table? P->(P ->(Q ^ P)). Now, does this mean that first, you work the Q^P? or do you work the p-> which is inside the ( ) first. Then do you take the result of whichever is used and connect it with p-> which is oustide the ( ). Do that make sense. I'm confused too. \n" );
document.write( "\r\n" );
document.write( "Start with this:\r\n" );
document.write( "\r\n" );
document.write( "P|Q| P->(P ->(Q ^ P))\r\n" );
document.write( "T|T| \r\n" );
document.write( "T|F| \r\n" );
document.write( "F|T| \r\n" );
document.write( "F|F| \r\n" );
document.write( "\r\n" );
document.write( "Now transfer TTFF under every P, and TFTF under every Q\r\n" );
document.write( "\r\n" );
document.write( "P|Q| P->(P ->(Q ^ P))\r\n" );
document.write( "T|T| T T T T \r\n" );
document.write( "T|F| T T F T \r\n" );
document.write( "F|T| F F T F \r\n" );
document.write( "F|F| F F F F\r\n" );
document.write( "\r\n" );
document.write( "Now take care of the ^, because that's the innermost\r\n" );
document.write( "operation. If ^ is between two T's it's T, otherwise it's F\r\n" );
document.write( "So we use this rule to fill in under the ^, but everytime\r\n" );
document.write( "we do we scratch through the letters on each side of it,\r\n" );
document.write( "like this:\r\n" );
document.write( "\r\n" );
document.write( "P|Q| P->(P ->(Q ^ P))\r\n" );
document.write( "T|T| T T T T T \r\n" );
document.write( "T|F| T T F F T \r\n" );
document.write( "F|T| F F T F F \r\n" );
document.write( "F|F| F F F F F\r\n" );
document.write( "\r\n" );
document.write( "Now I'll just erase them, but you can just leave them\r\n" );
document.write( "scratched:\r\n" );
document.write( "\r\n" );
document.write( "P|Q| P->(P ->(Q ^ P))\r\n" );
document.write( "T|T| T T T \r\n" );
document.write( "T|F| T T F \r\n" );
document.write( "F|T| F F F \r\n" );
document.write( "F|F| F F F\r\n" );
document.write( "\r\n" );
document.write( "Now take care of the second ->, because that's now the innermost\r\n" );
document.write( "operation. If -> has a T on the left and an F on the right,\r\n" );
document.write( "then it's F, otherwise it's T. So we use this rule to fill \r\n" );
document.write( "in under the ->, but as before, every time we do we scratch \r\n" );
document.write( "through the letters on each side of it, like this: \r\n" );
document.write( "\r\n" );
document.write( "P|Q| P->(P ->(Q ^ P))\r\n" );
document.write( "T|T| T T T T \r\n" );
document.write( "T|F| T T F F \r\n" );
document.write( "F|T| F F T F \r\n" );
document.write( "F|F| F F T F\r\n" );
document.write( "\r\n" );
document.write( "Again I'll erase the ones scratched, but you can just leave them\r\n" );
document.write( "scratched:\r\n" );
document.write( "\r\n" );
document.write( "P|Q| P->(P ->(Q ^ P))\r\n" );
document.write( "T|T| T T \r\n" );
document.write( "T|F| T F \r\n" );
document.write( "F|T| F T \r\n" );
document.write( "F|F| F T \r\n" );
document.write( "\r\n" );
document.write( "Finally we take care of the first ->, because that's now \r\n" );
document.write( "the innermost operation. We use the same rule again to fill \r\n" );
document.write( "in under the first ->,\r\n" );
document.write( "\r\n" );
document.write( "P|Q| P->(P ->(Q ^ P))\r\n" );
document.write( "T|T| T T T \r\n" );
document.write( "T|F| T F F \r\n" );
document.write( "F|T| F T T \r\n" );
document.write( "F|F| F T T \r\n" );
document.write( "\r\n" );
document.write( "And erasing those marked thru:\r\n" );
document.write( "\r\n" );
document.write( "P|Q| P->(P ->(Q ^ P))\r\n" );
document.write( "T|T| T \r\n" );
document.write( "T|F| F \r\n" );
document.write( "F|T| T \r\n" );
document.write( "F|F| T \r\n" );
document.write( "\r\n" );
document.write( "Edwin \n" );
document.write( " \n" );
document.write( " |