document.write( "Question 310536: Construct a truth table for p -> (~q ^ p) \n" ); document.write( "
Algebra.Com's Answer #222059 by Edwin McCravy(20055)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "Learn the rules:\r\n" ); document.write( "\r\n" ); document.write( " ~ always changes T to F and F to T, that is it always gives the opposite.\r\n" ); document.write( "\r\n" ); document.write( " V is usually T. The only times V is F is when there is an F on both sides of V\r\n" ); document.write( "\r\n" ); document.write( " ^ is usually F. The only times ^ is T is when there is a T on both sides of ^\r\n" ); document.write( "\r\n" ); document.write( "-> is usually T. The only times -> is F is when there is a T on the left of -> and a F on the right of ->\r\n" ); document.write( "\r\n" ); document.write( "That is, we learn the exceptional cases for V, ^ and ->. They are\r\n" ); document.write( "\r\n" ); document.write( "F V F becomes F, other cases T\r\n" ); document.write( "T ^ T becomes T, other cases F\r\n" ); document.write( "T -> F becomes F, other cases T \r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "p -> (~q ^ p)\r\n" ); document.write( "\r\n" ); document.write( "Start with this, so that you have all the \"pieces\", that is,\r\n" ); document.write( "You can build ~q from q. \r\n" ); document.write( "You can build (~q ^ p) from ~q and p.\r\n" ); document.write( "You can build p -> (~q ^ p) from p and (~q ^ p)\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( " | | | | |\r\n" ); document.write( " | | | | |\r\n" ); document.write( " | | | | |\r\n" ); document.write( " | | | | |\r\n" ); document.write( "\r\n" ); document.write( "Put TTFF under p and TFTF under q\r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\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 since q has TFTF under it, ~q must have the opposites,\r\n" ); document.write( "FTFT under it: \r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( "T | T | F | | |\r\n" ); document.write( "T | F | T | | |\r\n" ); document.write( "F | T | F | | |\r\n" ); document.write( "F | F | T | | |\r\n" ); document.write( "\r\n" ); document.write( "To fill in the next column, we use the rule for ^ which is\r\n" ); document.write( "\r\n" ); document.write( "Put an F unless there is a T\r\n" ); document.write( "on the left and a T on the right. Only in that one case do we put T\r\n" ); document.write( "\r\n" ); document.write( "In the top row ~q is F and p is T, so we must put F, because there is\r\n" ); document.write( "not a T on both sides of ^. This is not the exceptional case for ^.\r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( "T | T | F | F | |\r\n" ); document.write( "T | F | T | | |\r\n" ); document.write( "F | T | F | | |\r\n" ); document.write( "F | F | T | | |\r\n" ); document.write( "\r\n" ); document.write( "In the second row ~q is T and p is T, so we must put T, because there is\r\n" ); document.write( "a T on both sides of ^ and that is the one case of ^ when we must put a T.\r\n" ); document.write( " This is the exceptional case for ^.\r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( "T | T | F | F | |\r\n" ); document.write( "T | F | T | T | |\r\n" ); document.write( "F | T | F | | |\r\n" ); document.write( "F | F | T | | |\r\n" ); document.write( "\r\n" ); document.write( "In the third row ~q is F and p is F, so we must put F, because there is\r\n" ); document.write( "not a T on both sides of ^. This is not the exceptional case for ^.\r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( "T | T | F | F | |\r\n" ); document.write( "T | F | T | T | |\r\n" ); document.write( "F | T | F | F | |\r\n" ); document.write( "F | F | T | | |\r\n" ); document.write( "\r\n" ); document.write( "In the bottom row ~q is T and p is F, so we must put F, because there is\r\n" ); document.write( "not a T on both sides of ^. This is not the exceptional case for ^.\r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( "T | T | F | F | |\r\n" ); document.write( "T | F | T | T | |\r\n" ); document.write( "F | T | F | F | |\r\n" ); document.write( "F | F | T | F | |\r\n" ); document.write( "\r\n" ); document.write( "To fill in the last column, we use the rule for -> which is\r\n" ); document.write( "\r\n" ); document.write( "Put a T unless there is a T\r\n" ); document.write( "on the left and a F on the right. Only in that one case do we put F\r\n" ); document.write( "\r\n" ); document.write( "On the top line we put F because there is a T under p and an F under (~q ^ p),\r\n" ); document.write( "and that is the one case when we must put F under ->,\r\n" ); document.write( " This IS the exceptional case for ->. \r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( "T | T | F | F | F |\r\n" ); document.write( "T | F | T | T | |\r\n" ); document.write( "F | T | F | F | |\r\n" ); document.write( "F | F | T | F | |\r\n" ); document.write( "\r\n" ); document.write( "On the second line we put T because there is a T under p and a T under\r\n" ); document.write( "(~q ^ p). This is not the exceptional case for ->.\r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( "T | T | F | F | F |\r\n" ); document.write( "T | F | T | T | T |\r\n" ); document.write( "F | T | F | F | |\r\n" ); document.write( "F | F | T | F | |\r\n" ); document.write( "\r\n" ); document.write( "On the third line we put T because there is an F under p and\r\n" ); document.write( "an F under (~q ^ p). This is not the exceptional case for ->. \r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( "T | T | F | F | F |\r\n" ); document.write( "T | F | T | T | T |\r\n" ); document.write( "F | T | F | F | T |\r\n" ); document.write( "F | F | T | F | |\r\n" ); document.write( "\r\n" ); document.write( "Finally, on the fourth line we put T because there is an F under p and\r\n" ); document.write( "an F under (~q ^ p). This is not the exceptional case for ->.\r\n" ); document.write( "\r\n" ); document.write( "p | q | ~q | (~q ^ p) | p -> (~q ^ p) |\r\n" ); document.write( "--|---|----|----------|---------------|\r\n" ); document.write( "T | T | F | F | F |\r\n" ); document.write( "T | F | T | T | T |\r\n" ); document.write( "F | T | F | F | T |\r\n" ); document.write( "F | F | T | F | T |\r\n" ); document.write( "\r\n" ); document.write( "Edwin\n" ); document.write( " |