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How do i show that this statement is true?

$$\forall x,y,z,w \in \mathbb{Z}\space \space xSy \wedge zSw \implies (x+z)S(y+w)$$

The relation S is defined with:

$$ xSy \Leftrightarrow (x + y \textrm{ is even} \wedge x \leq y )$$

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What are $Z$ and $S$? – Michael Albanese Sep 29 '13 at 20:35
What does $S$ mean? [$Z$ usually means the integers, so are we doing arithmetic? But in arithmetic, $S$ is very often the successor relation -- and this statement is false.] – Peter Smith Sep 29 '13 at 20:37
Sorry for begin stupid :/ I added the definition of S. Z means the integers. – cenh Sep 29 '13 at 20:50
up vote 0 down vote accepted

You need to show that $(x+z)+(y+w)$ is even and $x+z\le y+w$, given that $x+y$ and $z+w$ are even, $x\le y,$ and $z\le w$. The first one should be simple, since $$(x+z)+(y+w)=x+z+y+w=x+y+z+w=(x+y)+(z+w).$$ For the second one, you'll start with $x+z$. Since $z\le w,$ then $x+z\le x+w$. What can you say about the order relationship between $x+w$ and $y+w$? What can you conclude, then, by transitivity of order?

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That x + w $\leq$ y + w I'm not entirely sure what I should conclude. – cenh Sep 29 '13 at 21:20
Yes, indeed. Since $x+z\le x+w$ and $x+w\le y+w,$ what can you conclude? – Cameron Buie Sep 29 '13 at 21:23
I'm not sure what I should conclude, do you mean something like x + z $\leq$ y + w – cenh Sep 29 '13 at 21:28
Exactly! That finishes the proof. – Cameron Buie Sep 29 '13 at 21:30
Thanks so much for taking the time to help me! – cenh Sep 29 '13 at 21:35



A and C => X 


B and D => Y


(A and C) and (B and D) => X and Y

More strong hint:

What do we have?

x + y is even and x ≤ y
z + w is even and z ≤ w

What should we get?

x + y + z + w is even and x + z ≤ y + w
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Could you explain a bit further, because I'm not exactly sure how this hint is helping me. – cenh Sep 29 '13 at 21:02
fixed first hint a bit, added 1 more – RiaD Sep 29 '13 at 21:11

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