# y is weakly preferred over x if and only if x+y ≤ 4 defines a preference relation on {0,1,2,3} why is this incomplete?

y is weakly preferred over x if and only if x+y ≤ 4 defines a preference relation on {0,1,2,3}. True or False?. I can see why it's not transitive but I was told it was incomplete if we take 2 and 3. If the bundle violates the condition x+y ≤ 4 (x ≤ 4-y) doesn't that just means x is weakly preferred over y hence complete.

Just like we deal with y is weakly preferred over x iff x≤y on the set X={0,1,2}.

Many thanks,

Define the Set of Alternatives X as X ={0,1,2,3}

Define the weakly preferred relation on X as

$$\succsim : X \to X$$ such that $$(\forall x \in X)(\forall y \in X)$$ , it is the case that

$$y \succsim x \iff x+y \leq 4$$

The definition for completeness can be referred below:- $$(\forall x \in X) (\forall y \in X) [ x \neq y \Rightarrow x \succsim y \: \lor y \succsim x] \Rightarrow \: \succsim$$ is complete

Clearly, we can find counter example to show that $$\succsim$$ is not complete.

Notice from the definition of $$\succsim$$, it also holds that $$\neg (y \succsim x) \iff \neg(x+y \leq 4)$$

Counterexample can be, $$\neg(3 \succsim 2)$$ as $$2+3 > 4$$ and $$\neg(2 \succsim 3)$$ as $$3+2>4$$

Since $$(\exists x \in X)(\exists y \in X) [x \neq y \land \neg (x \succsim y) \land \neg(y \succsim x)]$$ is true ,which is negation of definiton of completeness, it can be concluded that $$\succsim$$ is not complete.