Prove: (x ≤ y) → (x+z ≤ y+z)

  • Thread starter Firepanda
  • Start date
  • #1
Firepanda
430
0
[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:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
43,021
970
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
Firepanda
430
0
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
Firepanda
430
0
Basically I think I need to prove the lemma

x + z = y + z -> x = y
 

Suggested for: Prove: (x ≤ y) → (x+z ≤ y+z)

Replies
21
Views
901
Replies
12
Views
429
Replies
16
Views
445
Replies
15
Views
1K
Replies
23
Views
609
Replies
8
Views
368
Replies
1
Views
378
Replies
2
Views
582
Replies
32
Views
614
Top