document.write( "Question 1191577: please help me solve this problem: Construct a proof of the following theorem: [(P → Q)&(R → ¬Q)] →
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document.write( "¬(P&R) \n" );
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Algebra.Com's Answer #823370 by Edwin McCravy(20055)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "\r\n" ); document.write( "[(P → Q)&(R → ¬Q)] → ¬(P&R)\r\n" ); document.write( "\r\n" ); document.write( "Put TTTTFFFF under each P.\r\n" ); document.write( "Put TTFFTTFF under each Q.\r\n" ); document.write( "Put TFTFTFTF under each R.\r\n" ); document.write( "\r\n" ); document.write( "[(P → Q)&(R → ¬Q)] → ¬(P&R)\r\n" ); document.write( " T T T T T T \r\n" ); document.write( " T T F T T F\r\n" ); document.write( " T F T F T T\r\n" ); document.write( " T F F F T F\r\n" ); document.write( " F T T T F T\r\n" ); document.write( " F T F T F F\r\n" ); document.write( " F F T F F T\r\n" ); document.write( " F F F F F F\r\n" ); document.write( "\r\n" ); document.write( "Under the ¬ of ¬Q write the opposite of what's under\r\n" ); document.write( "the Q. Then erase what's under the Q\r\n" ); document.write( "\r\n" ); document.write( "[(P → Q)&(R → ¬Q)] → ¬(P&R)\r\n" ); document.write( " T T T F T T \r\n" ); document.write( " T T F F T F\r\n" ); document.write( " T F T T T T\r\n" ); document.write( " T F F T T F\r\n" ); document.write( " F T T F F T\r\n" ); document.write( " F T F F F F\r\n" ); document.write( " F F T T F T\r\n" ); document.write( " F F F T F F\r\n" ); document.write( "\r\n" ); document.write( "Under the → of P → Q write F for T → F\r\n" ); document.write( "and T for everything else. Then erase what's \r\n" ); document.write( "under P and Q:\r\n" ); document.write( "\r\n" ); document.write( "[(P → Q)&(R → ¬Q)] → ¬(P&R)\r\n" ); document.write( " T T F T T \r\n" ); document.write( " T F F T F\r\n" ); document.write( " F T T T T\r\n" ); document.write( " F F T T F\r\n" ); document.write( " T T F F T\r\n" ); document.write( " T F F F F\r\n" ); document.write( " T T T F T\r\n" ); document.write( " T F T F F\r\n" ); document.write( "\r\n" ); document.write( "Under the → of R → ¬Q write F for T → F\r\n" ); document.write( "and T for everything else. Then erase what's \r\n" ); document.write( "under R and ¬:\r\n" ); document.write( "\r\n" ); document.write( "[(P → Q)&(R → ¬Q)] → ¬(P&R)\r\n" ); document.write( " T F T T \r\n" ); document.write( " T T T F\r\n" ); document.write( " F T T T\r\n" ); document.write( " F T T F\r\n" ); document.write( " T F F T\r\n" ); document.write( " T T F F\r\n" ); document.write( " T T F T\r\n" ); document.write( " T T F F\r\n" ); document.write( "\r\n" ); document.write( "Under the & of P&R write T for T&T\r\n" ); document.write( "and F for everything else. Then erase what's \r\n" ); document.write( "under P and R:\r\n" ); document.write( "\r\n" ); document.write( "[(P → Q)&(R → ¬Q)] → ¬(P&R)\r\n" ); document.write( " T F T \r\n" ); document.write( " T T F\r\n" ); document.write( " F T F\r\n" ); document.write( " F T F\r\n" ); document.write( " T F F\r\n" ); document.write( " T T F\r\n" ); document.write( " T T F\r\n" ); document.write( " T T F\r\n" ); document.write( "\r\n" ); document.write( "Under the & of (P → Q)&(R → ¬Q) write T for T&T\r\n" ); document.write( "and F for everything else. Then erase what's \r\n" ); document.write( "under (P → Q) and (R → ¬Q):\r\n" ); document.write( "\r\n" ); document.write( "[(P → Q)&(R → ¬Q)] → ¬(P&R)\r\n" ); document.write( " F T \r\n" ); document.write( " T F\r\n" ); document.write( " F F\r\n" ); document.write( " F F\r\n" ); document.write( " F F\r\n" ); document.write( " T F\r\n" ); document.write( " T F\r\n" ); document.write( " T F\r\n" ); document.write( "\r\n" ); document.write( "Under the ¬ of ¬(P&R) write the opposite of what's under\r\n" ); document.write( "the &. Then erase what's under the &\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "[(P → Q)&(R → ¬Q)] → ¬(P&R)\r\n" ); document.write( " F F \r\n" ); document.write( " T T \r\n" ); document.write( " F T \r\n" ); document.write( " F T \r\n" ); document.write( " F T \r\n" ); document.write( " T T \r\n" ); document.write( " T T \r\n" ); document.write( " T T \r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Under the → of [(P → Q)&(R → ¬Q)] → ¬(P&R) write \r\n" ); document.write( "F for T → F and T for everything else. Then erase what's \r\n" ); document.write( "under [(P → Q)&(R → ¬Q)] and ¬(P&R):\r\n" ); document.write( "\r\n" ); document.write( "[(P → Q)&(R → ¬Q)] → ¬(P&R)\r\n" ); document.write( " T \r\n" ); document.write( " T \r\n" ); document.write( " T \r\n" ); document.write( " T \r\n" ); document.write( " T \r\n" ); document.write( " T \r\n" ); document.write( " T \r\n" ); document.write( " T \r\n" ); document.write( "\r\n" ); document.write( "Since we end up with all T's, the theorem is proved \r\n" ); document.write( "true in all 8 possible cases.\r\n" ); document.write( "\r\n" ); document.write( "Edwin\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |