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Prove: (x ≤ y) → (x+z ≤ y+z)

  1. Mar 17, 2012 #1
    [LOGIC] Prove: (x ≤ y) → (x+z ≤ y+z)

    I need to prove if x≤y then x+z ≤ y+z (for all x, y and z)

    Using these axioms (The first 17 are Tarski Arithmetic, and the following 7 are previously proved results)

    34zbrl1.png

    All I can think of so far is using Axiom TA16, but then what?

    Thanks
     
    Last edited: Mar 17, 2012
  2. jcsd
  3. Mar 17, 2012 #2

    HallsofIvy

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    Yes, that is one way to do it. Note that what you want to end with, [itex]x+z\le y+ z[/itex] is, again by TA16, equivalent to [itex]0\le (y+ z)- (x+ z)[/itex]. Do you see how to get to that?
     
  4. Mar 17, 2012 #3
    Thanks!

    Got that one now

    For another question on this example sheet I've almost done it apart from the last step where I have to show

    y + (-x) = 0 → y = x

    It seems so simple yet I can't think how to show that, any ideas?
     
  5. Mar 17, 2012 #4
    Basically I think I need to prove the lemma

    x + z = y + z -> x = y
     
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