document.write( "Question 1190422: MAT 145: Topics In Contemporary Math\r
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document.write( "QUESTION 7\r
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document.write( "Determine if the argument is valid or invalid using a truth table.\r
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document.write( "Premise: p rightwards arrow tilde q\r
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document.write( "Premise: p logical or q\r
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document.write( "Conclusion: p \n" );
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Algebra.Com's Answer #822065 by Edwin McCravy(20055)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "You can type a tilde for \"not\" or \"negation\". Look above the TAB key\r\n" ); document.write( "to the left of the 1 key. You have to hold the shift key down. ~\r\n" ); document.write( "You can make a right arrow for conditional\" or \"if then\" with a dash \r\n" ); document.write( "- and a greater than > to get ->. Use a little \"v\" for \"logical or\".\r\n" ); document.write( "For \"logical and\", you can use \"&\". \n" ); document.write( " Determine if the argument is valid or invalid using a truth table.\r \n" ); document.write( "\n" ); document.write( "Premise: p->~q \n" ); document.write( "Premise: pvq \n" ); document.write( "Conclusion: p \r\n" ); document.write( "\r\n" ); document.write( "Put & between the premises and -> before the conclusion\r\n" ); document.write( "\r\n" ); document.write( "[(p->~q)&(pvq)]->p\r\n" ); document.write( "\r\n" ); document.write( "You have to use parentheses and brackets so that there are only two\r\n" ); document.write( "things inside a parentheses or bracket with a v, &, or -> between \r\n" ); document.write( "them. \r\n" ); document.write( "\r\n" ); document.write( "Start with a column for each single letter\r\n" ); document.write( "for a statement on the left, and build it \r\n" ); document.write( "up piece by piece.\r\n" ); document.write( "Each heading must have either \r\n" ); document.write( "p\r\n" ); document.write( "q\r\n" ); document.write( "~p\r\n" ); document.write( "~q\r\n" ); document.write( "( ) v ( )\r\n" ); document.write( "( ) & ( )\r\n" ); document.write( "( ) -> ( )\r\n" ); document.write( "Single letters or letters with a tilde before them can \r\n" ); document.write( "replace the ( ), and the ( ) can have ~ before them.\r\n" ); document.write( "You can use brackets [ ] to enclose ( ). But remember\r\n" ); document.write( "each parentheses and each brackets can at most only \r\n" ); document.write( "contain two things with v, &, or -> between them. \r\n" ); document.write( "\r\n" ); document.write( "Start with this:\r\n" ); document.write( "\r\n" ); document.write( "| p | q | ~q | p->~q | pvq | (p->~q)&(pvq) || [(p->~q)&(pvq)]->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( "Put TTFF in the p column and TFTF in the q column:\r\n" ); document.write( "\r\n" ); document.write( "| p | q | ~q | p->~q | pvq | (p->~q)&(pvq) || [(p->~q)&(pvq)]->p |\r\n" ); document.write( "| T | T | | | | || | \r\n" ); document.write( "| T | F | | | | || | \r\n" ); document.write( "| F | T | | | | || | \r\n" ); document.write( "| F | T | | | | || |\r\n" ); document.write( "\r\n" ); document.write( "Next we have a ~q column. We put the opposite of what was in the q\r\n" ); document.write( "column:\r\n" ); document.write( "\r\n" ); document.write( "| p | q | ~q | p->~q | pvq | (p->~q)&(pvq) || [(p->~q)&(pvq)]->p |\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 | T | F | | | || |\r\n" ); document.write( "\r\n" ); document.write( "Next we have a p->~q column. The rule for -> is \r\n" ); document.write( "\"put T for everything except T->F, where there's a T on the \r\n" ); document.write( "left of -> and an F on the right of ->. That's the only time you \r\n" ); document.write( "put F.\" You have to look at the previous columns to see what's on \r\n" ); document.write( "the left of -> and what's on the right of it.\r\n" ); document.write( "\r\n" ); document.write( "| p | q | ~q | p->~q | pvq | (p->~q)&(pvq) || [(p->~q)&(pvq)]->p |\r\n" ); document.write( "| T | T | F | F | | || | \r\n" ); document.write( "| T | F | T | T | | || | \r\n" ); document.write( "| F | T | F | T | | || | \r\n" ); document.write( "| F | F | T | T | | || |\r\n" ); document.write( "\r\n" ); document.write( "Next we have a pvq column. The rule for v is \r\n" ); document.write( "\"put T for everything except FvF, where there's an F on both\r\n" ); document.write( "sides of the v. That's the only time you put F.\" You have to \r\n" ); document.write( "look at the previous columns to see what's on the left of v \r\n" ); document.write( "and what's on the right of it.\r\n" ); document.write( "\r\n" ); document.write( "| p | q | ~q | p->~q | pvq | (p->~q)&(pvq) || [(p->~q)&(pvq)]->p |\r\n" ); document.write( "| T | T | F | F | T | || | \r\n" ); document.write( "| T | F | T | T | T | || | \r\n" ); document.write( "| F | T | F | T | T | || | \r\n" ); document.write( "| F | F | T | T | F | || |\r\n" ); document.write( "\r\n" ); document.write( "Next we have a (p->~q)&(pvq) column. The rule for & is \r\n" ); document.write( "\"put F for everything except T&T, where there's an T on both\r\n" ); document.write( "sides of the &. That's the only time you put T.\" You have to \r\n" ); document.write( "look at the previous columns to see what's on the left of & \r\n" ); document.write( "and what's on the right of it.\r\n" ); document.write( "\r\n" ); document.write( "| p | q | ~q | p->~q | pvq | (p->~q)&(pvq) || [(p->~q)&(pvq)]->p |\r\n" ); document.write( "| T | T | F | F | T | F || | \r\n" ); document.write( "| T | F | T | T | T | T || | \r\n" ); document.write( "| F | T | F | T | T | T || | \r\n" ); document.write( "| F | F | T | T | F | F || |\r\n" ); document.write( "\r\n" ); document.write( "Finally we have a [(p->~q)&(pvq)]->p column. We already had a ->\r\n" ); document.write( "column. The rule, remember, for -> is \"put T for everything except \r\n" ); document.write( "T->F, where there's a T on the left of -> and an F on the right of\r\n" ); document.write( "->. That's the only time you put F.\" You have to look at the \r\n" ); document.write( "previous columns to see what's on the left of -> and what's on the \r\n" ); document.write( "right of it.\r\n" ); document.write( "\r\n" ); document.write( "| p | q | ~q | p->~q | pvq | (p->~q)&(pvq) || [(p->~q)&(pvq)]->p |\r\n" ); document.write( "| T | T | F | F | T | F || T | \r\n" ); document.write( "| T | F | T | T | T | T || T | \r\n" ); document.write( "| F | T | F | T | T | T || F | \r\n" ); document.write( "| F | F | T | T | F | F || T |\r\n" ); document.write( "\r\n" ); document.write( "Oh, oh! there's a F in that last column, so the argument is invalid.\r\n" ); document.write( "There must be only T's in the last column for an argument to be\r\n" ); document.write( "valid.\r\n" ); document.write( "\r\n" ); document.write( "Be sure to learn those rules for ~,v,->, and &. There is only one \r\n" ); document.write( "other, and that's equivalence <->. The rule for that is to put T if\r\n" ); document.write( "what are on both sides are the same (both T's or both F's), and F if \r\n" ); document.write( "they aren't the same.\r\n" ); document.write( "\r\n" ); document.write( "Edwin\n" ); document.write( " |