Question

U = Z
N(x): x is a non-negative integer
E(x): x is even
O(x): x is odd
P(x): x is prime
Translate into logical notation.
1. There exists an even integer.
2. Every integer is even or odd.
3. All prime integers are non-negative.
4. The only even prime is 2.
5. Not all integers are odd.
6. Not all primes are odd.
7. If an integer is not odd, then it is even.
solve any 1 or 2 parts in detail just to give me a headstart ......

Answer 1

2. \[\forall U \in (E(x) \nu O(x))\] is this correct?

Answer 2

1.
$$ \Large (\exists x \in \mathbb{Z}) ~~\mathbb{E}(x) $$
if you can assume the universe is Z then
$$ \Large \exists x ~\mathbb{E}(x) $$

2.
$$
\Large (\forall x \in \mathbb{Z} ) ~ ( \mathbb{E}(x) \lor \mathbb {O}(x))
$$

Answer 3

2. if we assume U = Z integers then we have
$$ \Large \forall x ~ ( \mathbb{E}(x) \lor \mathbb {O}(x)) $$

Answer 4

oh kky 3. \[(\forall x \in Z ) (\forall P(x) \in N(x))\]

Answer 5

3.
$$

\Large \forall x ~( \mathbb{P}(x) \to \mathbb{N}(x))


$$

Answer 6

where P(x) means x is prime, as you defined in your directions

Answer 7

and you defined the universe as Z, so we can assume that is the domain of discourse

Answer 8

\[what does this " \rightarrow " stands for?\]

Answer 9

"if P then Q "
$$ P \to Q $$

Answer 10

do you use a different symbol for 'if then' ?

Answer 11

no have read it in proposition logic only, didn't used it in predicate yet

Answer 12

predicate logic includes propositional logic. predicate logic is an extension of propositional logic

Answer 13

logical 'or' $$ \huge \lor $$ also comes from propositional logic

Answer 14

\[\leftarrow \rightarrow \] will be used for only then ? in 4.

Answer 15

the five propositional operations , 4 binary, 1 unary operator
$$
\Large ~\lor ~\land~ \to ~ \leftrightarrow ~ \neg $$

Answer 16

yes i can recall them now , ty

Answer 17

will 4 be something like this? or am i wrong again :/ ?(\[( E(2) \leftarrow \rightarrow P(x) )\]

Answer 18

almost

Answer 19

4.
$$
\Large \forall x ~ [(P(x) \land E(x)) \Rightarrow (x=2)]



$$

Answer 20

according to this there is a difference
$$ \Large \Rightarrow \\\rightarrow $$

http://en.wikipedia.org/wiki/Material_conditional

Answer 21

but we will use it as the simple propositional logic operator , logical conditional

Answer 22

just plain arrow

Answer 23

is it allowed to use propositional operator wit ha quantifier for e.g \[\Gamma \forall \]

Answer 24

we are only allowed to use the 5 simple operators as u listed

Answer 25

and $$ \Large \forall ~~ \exists $$ quantifiers

Answer 26

and set membership

Answer 27

yes exactly xD

Answer 28

$$
\Large x \in y $$
is acceptable .
But the way your homework is set up, we won't need to use set membership.
we will need to use equal sign though.
Your teacher defined E(x) as meaning $$ \Large x \in \mathbb{E} $$

Answer 29

5. \[\forall x (\Gamma x \in O(x) )\]

Answer 30

i hope i got it right this time .

Answer 31

$$

\Large \mathbb{E}(x) \\
\large \text{has the same meaning as} \\
\Large x \in \mathbb{E}
$$

Answer 32

oh kky didn't know that

Answer 33

5.
not all integers are odd.
means
' it is false that all integers are odd'

Answer 34

or better
' it is not the case that all integers are odd'
this is easiest to symbolize logically

$$
\Large \neg (\forall x ~\mathbb{O}(x))

$$

Answer 35

this is harder then i expected

Answer 36

$$
\Large \neg (\forall x ~\mathbb{O}(x))\\
\Large \equiv \exists x ~\neg \mathbb{O}(x)\\
\Large \equiv \exists x ~~\mathbb{E}(x)
$$

Answer 37

\[\Gamma (\forall P(x) \rightarrow O(x))\]

Answer 38

$$
\huge \equiv \\
\text{means logically equivalent, this is for the reader}
$$

Answer 39

it means just different ways of representing the same problem ? i can writer either one of the 3 and the answer will eb the same

Answer 40

right,

Answer 41

is the 6th one right ?

Answer 42

yes :)

Answer 43

your negation sign is a bit odd, capital gamma?

Answer 44

\neg is negation

Answer 45

yes xD couldn't find the negation sign so using gamma :p

Answer 46

7. \[\forall x \space (\neg (O(x)) \rightarrow E(x) )\]

Answer 47

$$
\Large \neg ~[~\forall x ~(\mathbb{P}(x)\to \mathbb{O}(x))]\\
\Large \equiv \exists x~\neg [~\mathbb{P}(x)\to \mathbb{O}(x)] \\
\Large \equiv \exists x~ (\mathbb{P}(x) \land \neg ~\mathbb{O}(x)) \\

$$

Answer 48

here i am using the rule
$$

\Large \neg (p \to q) \equiv p \land \neg~q

$$

Answer 49

useful to remember
$$
\Large p \to q \equiv \neg p \lor q\\


\Large \neg (p \to q) \equiv p \land \neg~q



$$

Answer 50

derived from implication definition then demorgan's law application right ?

Answer 51

right :)

Answer 52

is my 7 one right ?

Answer 53

yes just take out the parentheses around O(x)

Answer 54

$$
\large

\forall x \space (\neg O(x) \rightarrow E(x) )

$$

Answer 55

oh kky thankyou very very much for your help :)

Answer 56

this is just for reading, you did it correct

Answer 57

good job

Answer 58

thankyou :)