# Predicate Calculus help

• Mar 5th 2009, 10:08 AM
tokio
Predicate Calculus help
Use predicate calculus to display logical expressions for the following: The universe is ALL entities. You must use the notation that follows:

T(x): Denotes that x is a train
C(x): Denotes that x is a car
F(x,y): Denotes that x is faster that y
D(x,y): Denotes that x is driven by y
j: Denotes Jeff Gordon

1. All trains are faster than all cars.
2. Some trains are faster than some cars.
3. There is a car driven by Jeff Gordon that is faster than every train.
• Mar 5th 2009, 11:07 AM
Plato
Quote:

Originally Posted by tokio
There is a car driven by Jeff Gordon that is faster than every train.

$\left( {\exists x} \right)\left( {\forall y} \right)\left[ {C(x) \wedge D(x,j) \wedge \left[ {T(y) \Rightarrow F(x,y)} \right]} \right]$
• Mar 6th 2009, 09:56 AM
tokio
Thanks plato, i need help with teh first two and i cnat figure out how to use La-TEX to imput the universal instantiation and existential instantiation as well.

1. allxthereexistsyTCF(x,y) Any pointers on this one.
2. thereexistsxthereexistsy(T(x),C(y))

Im confused on the firsttwo because it seems like i have to change the universe for cars and im not 100% sure if this is logical correct.
• Mar 6th 2009, 11:58 AM
Predicate Logic
Hello tokio
Quote:

Originally Posted by tokio
...
T(x): Denotes that x is a train
C(x): Denotes that x is a car
F(x,y): Denotes that x is faster that y
...

1. All trains are faster than all cars.
2. Some trains are faster than some cars.

...

For (1), you need to say that if $x$ is a train and $y$ is a car, then $x$ is faster than $y$. Note that I have written this statement using singular words not plural. I've used phrases like ' $x$ is a train' and ' $y$ is a car'. This is simply because $x$ and $y$ can only stand for one thing at a time.

So, although we usually say that $\forall$ stands for the phrase 'For all ...', which is plural, it's more accurate to say that it stands for the phrase 'For each value of ...' which is singular.

Using this way of thinking, then, we need to translate into symbols:

For each value of $x$ and each value of $y$, $x$ is a train and $y$ is a car implies that $x$ is faster than $y$.

So this is:

$\forall x, y, [T(x) \wedge C(y)] \Rightarrow F(x, y)$

For number (2), although the words used are plural - some trains, some cars - again you need to think in the singular: think of $\exists$ as 'There exists a value of ...' - singular. So we re-write statement (2) as:

There exists a value of $x$ and a value of $y$, such that $x$ is a train and $y$ is a car and
$x$ is faster than $y$.

Can you do it now?

• Mar 6th 2009, 06:27 PM
tokio
Ok, i think i understand. So will 2 be this:

thereexistsanxthereexistsany[(T(x)&C(x)]impliesF(x,y)]
• Mar 6th 2009, 10:19 PM
Predicate Logic
Hello tokio
Quote:

Originally Posted by tokio
Ok, i think i understand. So will 2 be this:

thereexistsanxthereexistsany[(T(x)&C(x)]impliesF(x,y)]

No, not quite. The final connective is 'and', not 'implies'. So it reads:

$\exists x, y, T(x) \wedge C(y) \wedge F(x, y)$