CS1502, Spring 2009

Part I:====================
Which of the following are true, and which are false?

   A ^ B |=  A                  T
   A     |=  A ^ B              F
   A v B |=  B                  F
   B     |=  A v B              T
   ~(P ^ A) |= ~P v ~A          T
   ~P v ~A  |= ~(P ^ A)         T
   P |= Q                       F
   Q |= P                       F
   ~P |= P                      F
   P |= ~P                      F
   P ^ (~P v Q) |= Q            T

Which of the following are logical truths?
Which are logically satisfiable but not logical truths?
Which are not logically satisfiable?
(Note:  these involve the same sentences as above)

  (A ^ B) --> A  Logical truth

  A --> (A ^ B)  (logically satisfiable but not a logical truth)

    Not a logical truth:  It does not have to be true.
    E.g.:  woman(mary) --> woman(mary) ^ man(mary)
    clearly that could be false.
        
    is it satisfiable?  is there a world in which it is true?

    Sure.

    Suppose in our world, a is a cube and b is a tet.
    Let A be cube(a) and B be tet(b).

    cube(a) --> (cube(a) ^ tet(b)) -- that's true in our world.

  (A v B) -->  B   satisfiable but not a logical truth

     Not a logical truth:

     hockeyteam(penguins) v hockeyteam(steelers) -->
        hockeyteam(steelers) NOT true in our world.

    Is it satisfiable?  yes.

     A:  tet(a) B: cube(a)

     (tet(a) v cube(a)) -->  cube(a)

     If cube(a) is true in the world, then the
     sentence is true.

     It is also true if dodec(a) is true!

  B  -->  (A v B)  logical truth

  ~(P ^ A) -->  (~P v ~A) logical truth
  (~P v ~A)  --> ~(P ^ A) logical truth
  P --> Q   logically satisifiable but not LT
  Q --> P   logically satisfiable but not LT

  ~P --> P  

     P v P  equiv
     P

     The sentence is satisfiable -- it is true if P is true!

  P --> ~P   The sentence is satisfiable!

    ~P v ~P  equiv
    ~P

    The sentence is true if P is false.  
    (Contrast the to P |= ~P above!!!)

  (P ^ (~P v Q)) --> Q  logical truth

  Q --> (P ^ (~P v Q))  satisfiable but not a logical truth
     equivalent to:

     ~Q v (P ^ Q)
     (~Q v P) ^ (~Q v Q) 
     ~Q v P

     Clearly that may be true but does not have to be. 

  all x P(x) --> ~exists x ~P(x)  logical truth

  ~exists x ~P(x) --> all x P(x)  logical truth

  (P ^ (~P v Q)) --> Q (same as above by mistake)

  (A ^ B) <--> A     satisfiable, but not a logical truth
       (A ^ B) --> A     always true
       A --> A ^ B  could be true
                     
  A <--> (A ^ B)    not any different from previous sentence
       

  (A v B) <-->  B   satisfiable but not a logical truth
       (A v B) -->  B   satisfiable but not a logical truth
       B --> (A v B)   logical truth 

  B  <-->  (A v B)  this is the same as just above

  ~(P ^ A) <-->  (~P v ~A) logical truth
  (~P v ~A)  <--> ~(P ^ A) same

  P <--> Q  satisfiable not logical truth
  Q <--> P  satisfiable not logical truth

  Interesting!  Not satisfiable
  ~P <--> P   
    
      (~P --> P) ^ (P --> ~P)
      (P v P) ^ (~P v ~P)
      (P ^ ~P)
      cannot be true! 

  P <--> ~P  same

  (P ^ (~P v Q)) <--> Q  satisfiable but not logical truth

  Q <--> (P ^ (~P v Q)) same

  all x P(x) <--> ~exists x ~P(x) logical truth

  ~exists x ~P(x) <--> all x P(x) logical truth

  (P ^ (~P v Q)) <--> Q (repeated one by mistake)
 
Which of the following pairs are logically equivalent?
(Note1:  these involve the same sentences as above)

[These match the  are the <--> logical truths?]

  (A ^ B), A   no
  A, (A ^ B)   no
  (A v B),  B  no
  B ,  (A v B)  no
  ~(P ^ A),  (~P v ~A)  yes
  (~P v ~A) , ~(P ^ A)  yes
  P, Q   no
  Q, P   no
  ~P, P   no
  P, ~P   no
  (P ^ (~P v Q)), Q   no
  Q, (P ^ (~P v Q))   no
  all x P(x), ~exists x ~P(x)  yes
  ~exists x ~P(x), all x P(x)  yes
  (P ^ (~P v Q)), Q   no

Which of the following are logically valid arguments?
(Note:  these involve the same sentences as above)

[These match the |= sentences above]

  (A ^ B)
  ---
   A    valid

  A
 ---
 (A ^ B)  nope

  (A v B)
  ---
  B       nope

  B 
  ---
 (A v B)  valid

  ~(P ^ A)
   ---
  (~P v ~A)   valid

  (~P v ~A) 
   ---
   ~(P ^ A)  valid

  P
  ---
  Q     nope

  Q
  ---
  P     nope

  ~P
  --
  P    nope

  P
  ---
  ~P     nope

  (P ^ (~P v Q))
  ---
  Q      valid

  Q
  ---
 (P ^ (~P v Q))  nope

  all x P(x)
  ---
 ~exists x ~P(x)  valid

  ~exists x ~P(x)
  ---
  all x P(x)    valid

  (P ^ (~P v Q))
  ---
   Q      valid

Part II:=============================
For each of the following, state whether it is a valid argument.
If it is not valid, show that it is not valid.

all x (Student(x) --> Smart(x))
all x Student(x)
---
all x Smart(x)

valid
--------------------------
all x Student(x)
all x Smart(x)
---
all x (Student(x) ^ Smart(x))

valid
--------------------------
exists x (Student(x) --> Smart(x))
exists x Student(x)
---
exists x Smart(x)

no
--------------------------
exists x Student(x)
exists x Smart(x)
---
exists x (Student(x) ^ Smart(x))

no
--------------------------
--------------------------
Are the following pairs logically equivalent?
If not, show that they are not.

all x (P(x) ^ Q(x)),  
all x P(x) ^ all x Q(x)

Yes
--------------------------
all x (P(x) v Q(x)),
all x P(x) v all x Q(x).

no
--------------------------
exists x (P(x) v Q(x)),
exists x P(x) v exists x Q(x)

yes
--------------------------
exists x (P(x) ^ Q(x)),
exists x P(x) ^ exists x Q(x)

Ans: no (goes ==> but not <==)  

Part III:=============================

Consider a world with only two shapes.  They are
both cubes and they are in the same row. Is the following
sentence true? [Note: this question was revised 3-17-09; it originally
said "both tets", a cut-and-paste mistake]

all x all y 
((cube(x) ^ cube(y)) --> (leftof(x,y) v rightof(x,y)))

Not if we assume that a shape cannot be leftof or rightof itself.

As an aside:  how can we say that a shape cannot be leftof or rightof
itself?  

all x,y (leftof(x,y) --> ~(x = y))  or, equivalently,
all x,y (x = y --> ~leftof(x,y))


----------------------------
Express the following sentences in English
(where taken(x,y) means that student x took class y):

exists x exists y taken(x,y)
There is a student who took a class

exists x all y taken(x,y)
There is a student who has taken all classes

all x exists y taken(x,y)

All students have taken at least one class
(but could be different classes by different students)

exists y all x taken(x,y)
There is a class that all students have taken

all y exists x taken(x,y)
for all cs classes, there is at least one student who took it
(but it can be different for different students)

exists y (small(y) ^  all x(small(y) --> y=x))
There is exactly one small thing in the world.
-------------

Suppose Q(x,y,z) is the statement x+y=z
The domain of discourse is the real numbers.

Are the following sentences true?

all x all y exists z Q(x,y,z)
True.

exists z all x all y Q(x,y,z)
False.

---------
Express the following sentences in logic
(where L(x,y) means that x loves y, and the domain of 
discourse for both x and y is the set of all people 
in the world)

everybody loves fred
all x L(x,fred)

every loves somebody
all x exists y L(x,y)

there is someone whom no one loves
exists x all y ~L(y,x) [changed 3-17-09:  the y and x were backwards]

everyone loves himself or herself
all x L(x,x)

--------------------
Express the following in logic. First do this using
both function and predicate symbols, and then do it using only
predicate symbols.

Everyone has exactly one mother, who is a woman.  

Domain of discourse:  people

Assuming "mother" is defined for each element in the domain of discourse:

all x Olderthan(mother(x),x)

Here is the version with predicate symbols.

all x exists y (Motherof(y,x) ^ Olderthan(y,x) ^ 
                all z(MotherOf(z,x) --> y=z))