Prove: (x ≤ y) → (x+z ≤ y+z)

  • Thread starter Firepanda
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  • #1
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[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
 
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Answers and Replies

  • #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?
 
  • #3
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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?
 
  • #4
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Basically I think I need to prove the lemma

x + z = y + z -> x = y
 

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