Results 1 to 4 of 4

Math Help - Show that each conditional statement is a tautology without using truth tables.

  1. #1
    Junior Member
    Joined
    Apr 2013
    From
    USA
    Posts
    69
    Thanks
    1

    Show that each conditional statement is a tautology without using truth tables.

    1. [(p-->q) ^ (q-->r)] --> (p-->r)

    2. [p ^ (p-->q)]-->q


    My book isn't exactly making it clear on how to do problems like these. Very frustrating. Can someone lead me in the right direction? All I know is that I'm supposed to use Logical Equivalences.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1

    Re: Show that each conditional statement is a tautology without using truth tables.

    Quote Originally Posted by lamentofking View Post
    2. [p ^ (p-->q)]-->q.
    \begin{array}{*{20}{c}}{\left[ {p \wedge \left( {p \to q} \right)} \right]}\\{\left[ {p \wedge \left( {\neg p \vee q} \right)} \right]}\\{\left( {p \wedge \neg p} \right) \vee \left( {p \wedge q} \right)}\\{\left( {p \wedge q} \right)}\\q\end{array}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2013
    From
    USA
    Posts
    69
    Thanks
    1

    Re: Show that each conditional statement is a tautology without using truth tables.

    Quote Originally Posted by Plato View Post
    \begin{array}{*{20}{c}}{\left[ {p \wedge \left( {p \to q} \right)} \right]}\\{\left[ {p \wedge \left( {\neg p \vee q} \right)} \right]}\\{\left( {p \wedge \neg p} \right) \vee \left( {p \wedge q} \right)}\\{\left( {p \wedge q} \right)}\\q\end{array}
    Thanks. So would the first one be:

    [(p-->q) ^ (q-->r)] --> (p-->r)

    [(~p v q) ^ (~p v r)]
    (~p v q ^ ~p) ^(~p v q v r)
    .....

    Okay I'm stuck. lol. I guess I missed the first step huh. I don't know what I'm doing wrong.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,528
    Thanks
    773

    Re: Show that each conditional statement is a tautology without using truth tables.

    For the hypothetical syllogism (as opposed to categorical syllogism), it may be easier to prove that the negation is false. Note that ~(p -> q) = p /\ ~q. Take the negation of the whole formula and apply this rule to the outer implication (third from the left) and to p -> r. Replace the first two implications with disjunctions. Then use distributivity.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 16th 2013, 03:27 PM
  2. Prove a tautology without truth tables.
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: September 7th 2010, 06:37 AM
  3. Replies: 3
    Last Post: January 21st 2010, 07:45 AM
  4. Replies: 1
    Last Post: March 27th 2009, 06:43 AM
  5. Tautology, truth table
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: February 23rd 2009, 07:12 PM

Search Tags


/mathhelpforum @mathhelpforum