Results 1 to 1 of 1

Thread: Proving the Hypothetical Syllogism using Induction

  1. #1
    Sep 2009

    Proving the Hypothetical Syllogism using Induction

    Hi All,

    First post here. thanks for the help in advance!

    Okay first, the problem:

    In Logic, the hypothetical syllogism says that given three statements, P, Q, and R, the hypothesis P $\displaystyle \rightarrow$ Q and Q $\displaystyle \rightarrow$ R logically imply P $\displaystyle \rightarrow$ R. Symbolically written ((P $\displaystyle \rightarrow$ Q) ^ (Q $\displaystyle \rightarrow$ R) $\displaystyle \Rightarrow$ (P $\displaystyle \rightarrow$ R))
    Use induction to show that, given n statements ($\displaystyle ^A{1}$, $\displaystyle ^A{2}$, .... $\displaystyle ^A{n}$) that ($\displaystyle ^A{1}$ $\displaystyle \rightarrow$ $\displaystyle ^A{2}$) ^ ($\displaystyle ^A{2}$ $\displaystyle \rightarrow$ $\displaystyle ^A{3}$ ^ ... ^ ($\displaystyle ^A{n-1}$ $\displaystyle \rightarrow$ $\displaystyle ^A{n}$) $\displaystyle \Rightarrow$ ($\displaystyle ^A{1}$ $\displaystyle \rightarrow$ $\displaystyle ^A{n}$) for n $\displaystyle \geq$ 2

    The problem I am having is that I am not sure how to work out an equation from this statement that I can start to use induction on. Also, I'm not sure if I should try to prove this using an inequality expression or an = expression. It would seem to me that an inequality would make more sense in an implication of this sort, but I am anything but positive on it.

    I tried subbing in an arbitrary x for $\displaystyle ^A{1}$, and assigning (x+1) to $\displaystyle ^A{2}$, and using $\displaystyle ^A{1}$ $\displaystyle \leq$ $\displaystyle ^A{2}$, but my problem exploded and became unreasonable.

    Can someone please help me find what recursive function or sum equation I should start with? I should be able to solve the proof from there with mathematical induction (also, do you suppose strong induction or standard will work in this case?)

    Thanks again all!


    Here is what I have done so far. I think it proves this is true for $\displaystyle ^A{n}$ when f(n)=n, but is this all I need to do? What about cases where f(n) is equal to something other than just n?

    let $\displaystyle ^A{n}$ = f(n) where:
    f(1) = 1
    f(n) = n for n $\displaystyle \geq$ 2

    Step 1: (Base case) LHS $\displaystyle ^A{2}$ = 2, and RHS f(2) = 2
    Step 2: (Show $\displaystyle ^A{n+1}$ = f(n+1)
    $\displaystyle ^A{n+1}$ = f(n+1)
    $\displaystyle ^A{n+1}$ = n+1 $\displaystyle \leftarrow$ induction step

    so there is the proof for $\displaystyle ^A{n+1}$, but something just feels wrong about this. have I solved this problem, or am I missing something here?
    Last edited by TravisH82; Sep 30th 2009 at 08:45 AM. Reason: Showing my work so far
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fun statistical problem inside a hypothetical situation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Nov 2nd 2009, 06:47 PM
  2. Carroll syllogism and sets
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: Oct 5th 2009, 06:18 AM
  3. proving (by induction)
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Aug 30th 2009, 04:50 PM
  4. Hypothetical integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 18th 2009, 05:18 PM
  5. Replies: 15
    Last Post: Jun 3rd 2008, 03:58 PM

Search tags for this page

Search Tags

/mathhelpforum @mathhelpforum