Assertion of Type Inference Rules/Type Checking
Well, you need to construct a typing derivation which has what you want to prove as a conclusion, and no open premise (i.e., leaves must be axioms, or rules with no premises). In the case of your example, the conclusion will be

$$ int\;x,\,int\;y\;\vdash\;(x &lt; (x \times y) + x) : bool $$

so the proof will start by

$$ {int\;x,\,int\;y\;\vdash\;x : int \quad\quad\quad
  int\;x,\,int\;y\;\vdash\;((x \times y) + x) : int}
  \over {int\;x,\,int\;y\;\vdash\;(x &lt; ((x \times y) + x)) : bool} $$

and then you continue building up the tree until you reach leaves (i.e., rules with no premises). This will immediately be the case in the left branch since you only have a variable.



Here is the full proof (as an image):