Logic Proof question?

To prove W is a partial order, we need to show transitivity, reflexivity, and antisymmetry. We'll just split them up and do each part separately.

Part 1: Reflexivity: xWx for all x.

Consider any x in R. Then x=x+7a for a=0 in Q. Thus xWx.

Part 2: Transitivity: xWy and yWz implies xWz for all such x,y,z.

Assume xWy and yWz, so that for a,b≥0 in Q we know:

y = x+7a

z = y+7b

Then we know z = y+7b = (x+7a) + 7b = x + 7(a+b). Since a and b are in Q, a+b is in Q. Since a,b≥0, we know a+b≥0. This implies xWz.

Part 3: Antisymmetry: If xWy and yWx, then x=y.

Assume we have xWy and yWx. Then there are a,b≥0 in Q so that:

x = y + 7a

y = x + 7b

We can just rely on our old order on R to prove this. Just look:

x = y + 7a ≥ y

y = x + 7b ≥ x

And since the standard order ≥ is a partial (indeed, a total) order on R, we know x=y from the antisymmetry of ≥. Thus x=y.

Given parts 1,2,3, we know W is a partial order. QED