1. ## quantifier

T(x) : x is in this class
B(x) : x has a motor bike
C(x) : x has a car

how do I state:
Only one person in this class has a motor bike

my current answer is:

$\exists x\forall y[(x \neq y) \wedge T(x) \wedge B(x)]$

is this right?

how about if there is at least two people in this class have a car

$\exists x\exists y[T(x) \wedge T(y) \wedge C(x) \wedge C(y)]$

2. Originally Posted by TheRekz
T(x) : x is in this class
B(x) : x has a motor bike
C(x) : x has a car
Only one person in this class has a motor bike
There is at least two people in this class have a car
For the first on I would expect:
$\left( {\exists x} \right)\left( {\forall y} \right)\left[ {\left( {T(x) \wedge B(x)} \right) \wedge \left( {\left( {T(y) \wedge B(y)} \right) \Rightarrow \left( {x = y} \right)} \right)} \right]$.

The second one could be:
$\left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {T(x) \wedge C(x)} \right) \wedge \left( {\left( {T(y) \wedge C(y)} \right) \wedge \left( {x \not= y} \right)} \right)} \right]$

3. Originally Posted by Plato
For the first on I would expect:
$\left( {\exists x} \right)\left( {\forall y} \right)\left[ {\left( {T(x) \wedge B(x)} \right) \wedge \left( {\left( {T(y) \wedge B(y)} \right) \Rightarrow \left( {x = y} \right)} \right)} \right]$.

The second one could be:
$\left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {T(x) \wedge C(x)} \right) \wedge \left( {\left( {T(y) \wedge C(y)} \right) \wedge \left( {x \not= y} \right)} \right)} \right]$
for the first one would it be the same if I say y = x, rather than x = y??

4. WELL, of course!
It is always true that $\left( {x = y} \right) \equiv \left( {y = x} \right)$

5. Originally Posted by Plato
WELL, of course!
It is always true that $\left( {x = y} \right) \equiv \left( {y = x} \right)$
thanks Plato, it really helps.. what about the second one?? is my answer considered fine?

6. Look at the answer I gave for the second one:
$\left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {T(x) \wedge C(x)} \right) \wedge \left( {\left( {T(y) \wedge C(y)} \right) \wedge \left( {x \not= y} \right)} \right)} \right]$

7. Originally Posted by Plato
Look at the answer I gave for the second one:
$\left( {\exists x} \right)\left( {\exists y} \right)\left[ {\left( {T(x) \wedge C(x)} \right) \wedge \left( {\left( {T(y) \wedge C(y)} \right) \wedge \left( {x \not= y} \right)} \right)} \right]$
yeah but I think it's almost similar from what I did

8. But in what you did it is possible that x=y.
In that case we don't have two, but maybe only one.