Critique my proof involving quantifiers

[Baron]

Prove or disprove: For all x, y ∈ R (real number), if x < y + z for every z > 0, then x < y

My work:

Disprove by providing a counterexample.

If x = 1, y = 0 and z = 2 which is z > 0, then x < y + z as 1 < 0 + 2 but 1 ≮ 0

 

[pka]

Prove or disprove: For all x, y ∈ R (real number), if x < y + z for every z > 0, then x < y
Disprove by providing a counterexample.
If x = 1, y = 0 and z = 2 which is z > 0, then x < y + z as 1 < 0 + 2 but 1 ≮ 0

That is not a counterexample. It is not true \(\displaystyle \forall z>0\). Take \(\displaystyle z=0.5\).

However, it \(\displaystyle x=0~\&~y=0\) then \(\displaystyle (\forall z>0)[x<y+z]\).

 

[Baron]

That is not a counterexample. It is not true \(\displaystyle \forall z>0\). Take \(\displaystyle z=0.5\).

However, it \(\displaystyle x=0~\&~y=0\) then \(\displaystyle (\forall z>0)[x<y+z]\).

That makes sense.

Did I accidentally disprove the statement: For all z > 0 and for all x, y ∈ R (real number), if x < y + z, then x < y.

 

[pka]

That makes sense.

Did I accidentally disprove the statement: For all z > 0 and for all x, y ∈ R (real number), if x < y + z, then x < y.

But this is true: If \(\displaystyle \{x,y\}\subset\mathbb{R}~\&~(\forall z>0)[x<y+z]\) implies \(\displaystyle x\le y\).