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Should've paid attention in math foundations. Can anyone solve this using direct, contradiction or contrapositive?

$\forall x, y \in \mathbb{R}$, if $x \lt y$, then $2 + x \lt 4 + y$.

This is the work I've done so far,

$x \lt y$

$\Rightarrow x + 1 \lt y + 1$ (O3)

$\Rightarrow x + 1 + 1 \lt y + 1 + 1$ (O3)

$\Rightarrow x + 2 \lt y + 2$

And I've also proved that $2 \lt 4$, but I cannot figure out how to properly connect this to get the solution of $x + 2 \lt y + 4$.

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    foundations ??? It's elementary algebra. – 2017-01-25
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    With "O3", you seem to be referring to properties of the order relation you are allowed to use. It would be best to include the list of properties you are allowed to use, or give a reference to them. – 2017-01-25
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    (To the OP) This question would be a good candidate for reopening if you would provide the axioms as indicated by @StackTD. Also, see my edit and and try to take full advantage of [MathJax](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) for your math formatting in the future. :-) – 2018-08-30

1 Answers 1

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Hint: $x

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    Appreciate the hint, I should've shown the work I had so far which was this. Can't figure out how to get past that. – 2017-01-25