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Conditionals


IF —> THEN


   

  P Q  P —> Q
0) 0 0     1
1) 0 1     1
2) 1 0     0
3) 1 1     1



Assume the instructor promises his students that
if they maintain perfect attendance they will pass the course.

Let
A = You have perfect attendance.
B = You will pass the course.

Let A —> B mean
"If you have perfect attendance, then you will pass the course."

0) Assume that you have imperfect attendance and you fail the course.
Is the statement A —> B a lie? No. The promise was not broken.
Since the implication is not false, we must say then that it is true.
In our true or false world, there is nothing in between true and false.
The opposite of false is always true.
And a statement cannot be both true and false.

1) Assume that you do not maintain perfect attendance and you pass the course.
Is the statement A —> B a lie? No. The promise is irrelevant (i.e. moot).
Since the implication is not false, we must say it is true.

2) Assume that you do maintain perfect attendance and you fail the course.
Is the implication A —> B a lie? Yes. The promise was broken.
The implication is false; we cannot say it is true.

3) Assume that you do maintain perfect attendance and you pass the course.
Is the statement A —> B a lie? No. The promise is fulfilled.
The implication is obviously true.


With these 4 conditions, we can now make a truth table for any IF statement ->

    A B |  A -> B
    _____________

0)  0 0 |    1
1)  0 1 |    1
2)  1 0 |    0
3)  1 1 |    1
    _____________

The IF table has at least 8 translations. Each means the same thing.

A implies B.
If A is true, then B is true.
A only if B.
B if A.
A is sufficient for B.
B is necessary for A.
All A's are B's.
Either A is false or B is true.

ASSIGNMENT


Choose two compound statements from the list below.
Check the comments below first.
Each of your statements may have already been chosen,
but do not choose a pair already chosen.
If someone has previously chosen 1 and 2,
you may choose 2 and 5, and someone else may choose 1 and 5.
     
01.  –A + B

02.   A + –B

03.  –A & B

04.   A & –B

05.  –A + –B

06.  –A & –B

07.  –(A + B)

08.  –(A + –B)

09.  –(A & B)

10.  –(A & –B)

11.  –(–A + –B)

12.  –(–A & –B)

Make truth tables for both compound statements.
Call the first compound statement P, and the second one Q.
Now make a third truth table for P->Q,
and a fourth for Q->P.
Q->P is called the converse of P->Q.

Create two simple statements A and B
that no one else has previously placed on the wiki.
Translate P->Q and Q->P into compound English sentences.

Put your 2 conditional statements and their accompanying truth tables into one comment below;
and mind your P's and Q's.
If a neighbor has made an error, help them out.

There are more than 3 million truth table web sites listed by Google .
Here is one I found when I searched for Truth Tables.
You may find it clearer than my explanations.

Put your 2 conditional statements and their accompanying truth tables into one comment below;and mind your P's and Q's.If a neighbor has made an error, help them out.There are more than 3 million truth table web sites listed by .Here is one I found when I searched for .You may find it clearer than my explanations.


EXAMPLE:
I am choosing statements 3 and 5 for my example.
This means you cannot choose 3 and 5.
You may, of course, choose 3 or 5, but not both.


3.  A B |  –A & B  = P              5.  A B | –A + –B = Q
    ___________________                 __________________
                                                            
0)  0 0 |  1    0  | 0                  0 0 | 1    1  | 1
1)  0 1 |  1    1  | 1                  0 1 | 1    0  | 1
2)  1 0 |  0    0  | 0                  1 0 | 0    1  | 1
3)  1 1 |  0    1  | 0                  1 1 | 0    0  | 0
    ___________________                 __________________

Notice that I negated the A and B where indicated
before I compounded with & or + using the AND and OR tables.
The results are in columns P and Q.
Always follow this order of operations from first to last: () — & + —>
This means compound within the parentheses first,
then negate before ANDing, AND before ORing, and do the IFs last.

Does #3 imply #5? Let's see if P->Q.
Copy columns P and Q from the tables above to the table below.

    P Q |  P -> Q
    ______________

0)  0 1 |    1  
1)  1 1 |    1  
2)  0 1 |    1  
3)  0 0 |    1  
    ______________

Next refer to the IF table at the top of the page for the values of P -> Q .
Notice that if P, the premise, is false, then the implication -> is always true,
no matter what the value of Q, the conclusion.
Therefore, the implication -> is only false when P is true and Q is false.
In rows 0, 1, 2, 3 of this example, a true is never followed by a false.
So my P -> Q is always true under all conditions.
A statement that is True under all conditions is called a TAUTOLOGY.
Since P = –A & B,
and Q = –A + –B,
(–A & B) -> (–A + –B) is a tautology.

Is Q -> P also a tautology? Does #5 imply #3 ? Let's see if Q -> P .

    Q  P |  Q -> P
    ______________

0)  1  0 |    0  
1)  1  1 |    1  
2)  1  0 |    0  
3)  0  0 |    1  
    ______________

So Q -> P is not a tautology.
But notice that the outcome in the right column
is the same as column B in all two statement tables.
Therefore, (–A + –B) -> (–A & B) = B.
That's a hard way to say something so simple.

After you have created your four tables,
create two simple statements for A and B that no one else created.

Here are my two simple statements:

A = Mr. Allman shaved himself.
B = The Barber shaved Mr. Allman.

Now translate P -> Q and Q -> P into English:

P -> Q = (–A & B) -> (–A + –B) =
If Mr. Allman did not shave himself and the Barber shaved him,
then one or the other did not shave him.

Q -> P = (–A + –B) -> (–A & B) =
If Mr. Allman did not shave himself, or the Barber did not shave him,
then Mr. Allman did not shave himself and the Barber shaved him instead.

As we have seen, this convoluted statement is not a tautology,
but it is equivalent to B, The Barber shaved Mr. Allman.
Therefore Q -> P is true if and only if B is true.

You earn 7 points for this assignment:
4 points: 1 for each of the four tables,
1 point for the simple statements A and B, and
2 points: 1 for each of the two English translations.

Click her for a Conditional IF Demo.

HELP!
To speed up your exercise:
Hilite, copy, and paste the two blank tables below into your comment.
Replace the ? columns with your two choices from above.
To hilite, press left mouse key and drag mouse across table.
To copy any hilited text, press CTRL+C, or CTRL+INSERT.
To paste, press CTRL+V, or SHIFT+INSERT.
Add and delete columns as necessary.
                                      

 A B?=P    AB ?=Q
 0 0  

 

   0 0  
 0 1       0 1  
 1 0       1 0  
 1 1       1 1  

                                   

 PQP -> Q   QPQ -> P
             
             
             
             

Comments:

From wHolt - 12/10/06 9:44 AM

Bubba -

1 + anything = 1
0 + 0 = 0

From Bubba - 12/9/06 7:31 AM

ABA+-B   AB-A&B 
 0 0  0  1   0   00  1   O 0
 0 10    0      0 0  1 1   1 1
 1 0 1    10  1 0 0   0 0
 1 1   1  0 1  1 1 0   1 0
PQP -> Q T  QPQ -> P
 1 0      0  0 0 0 
 0 1      1  0 1 1 
 1 0      0  0 0 0 
 1 0      0  0 0 0 

NUMBERS 2&3

A = you are good at sports.

B = you will get a scholarship.

From wHolt - 12/7/06 11:07 AM

Bubba - Your A+-B table is not correct. You did A&~B instead. Review.

From Bubba - 12/6/06 10:17 AM

ABA+-B   AB-A&B 
 0 0  0  1   0   00  1   O 0
 0 10    0      0 0  1 1   1 1
 1 0 1    1 1  1 0 0   0 0
 1 1   1  0 0  1 1 0   1 0
PQP -> Q T  QPQ -> P
 1 0      0  0 0 0 
 0 1      1  0 1 1 
 1 0      0  0 0 0 
 1 0      0  0 0 0 

NUMBERS 2&3

A = you are good at sports.

B = you will get a scholarship.

From wHolt - 10/24/06 11:08 AM

Spartan -
test your P and Q in the applet at the top of the page.
click the button to start it.

From Spartan - 10/23/06 3:12 PM

7. -(A+B)

9. -(A&B)

A= I stayed up late.

B= I am tired.

A B-(A+B)=P    AB -(A&B)=Q
 0 0 1     11 

 

   0 0 1    11 
 0 1 1     0 1      0 1 1    00 
 1 0 0     11     1 0 0    10 
 1 1 0     00      1 1 0    0 0 

                                   

 PQP -> Q   QPQ -> P
 11     1    1 1  1
 10    1    0 1  0
 10     1    0 1  0
 00     1    0 0  1

From wHolt - 10/20/06 1:21 PM

Tiger - Thanks, for finishing yours.

From Tiger - 10/19/06 5:19 PM

 A B-A&B=P    AB -(A+-B)=Q
 0 0 1   0  0

 

   0 0 0     1  0
 0 1 1    1  1    0 1 0     0  1
 1 00     0  0    1 0 1      1  0
 1 1 0    1  0    1 1 1      0  0

                                   

 PQP -> Q   QPQ -> P
 0 0    1     0 0    1
 1 1     1    1 1     1
 0 0     1    0 0     1
 0 0     1    0 0     1

#3 and #8

A=I like apples.

B= I like oranges.

P->Q=(-A&B)->-(A+-B)=If I do not like apples and I like oranges, then I do not like apples or I like oranges.

Q->P=-(A+-B)->(-A&B)=If I do not like apples or I like oranges, then I do not like apples and I like oranges.

From Fro - 10/18/06 7:11 PM

A

B

     #1 
-A  + B

=P
  A
B 
     #7
-(A +B)
=Q
 0
 0
  1    0
 1
 
 0
 0
  1    1
 1
 0
 1
  1    1
 1
 
 0
 1
  1    0
 0
 1
 0
  0    0
 0 
 1
 0
  0    1
 0
 1
 1
  0    1
 1 
 1
 1
  0    0
 0
PQP -> Q   QPQ -> P
 1 1   1     11   1
 1 0   1     01   1
 0 0   0     00   0
 1 0   1     01   1

A: I can get this one.

B: I move on.

P→Q=(-A+B)→ -(A+B)

If I can NOT get this one or I move on THEN I can NOT get this one NOR will I move on.

Q→P= -(A+B)→(-A+B)

If I can NOT get this one NOR move on THEN I can NOT get this one or I move on..............

From wHolt - 10/18/06 11:15 AM

Fro- Q is still wrong
You did not use Brian's applet.
Cant make you.
So...
Do a table for A+B.
Then negate the results.
It must be hard to do.

From Fro - 10/17/06 6:45 PM

A

B

     #1 
-A  + B

=P
  A
B 
     #7
-(A +B)
=Q
 0
 0
  1    0
 1
 
 0
 0
  1    1
 0
 0
 1
  1    1
 1
 
 0
 1
  1    0
 1
 1
 0
  0    0
 0 
 1
 0
  0    1
 1
 1
 1
  0    1
 1 
 1
 1
  0    0
 1
PQP -> Q   QPQ -> P
 1 0   0     01   1
 1 1   1     11   1
 0 1   1     10   0
 1 1   1     11   1

A: I can get this one.

B: I move on.

P→Q=(-A+B)→ -(A+B)

If I can NOT get this one or I move on THEN I can NOT get this one NOR will I move on.

Q→P= -(A+B)→(-A+B)

If I can NOT get this one NOR move on THEN I can NOT get this one or I move on.

From wHolt - 10/17/06 2:27 PM

Those below have made it past the conditional dragon —
If you are not on the list yet, go slower so you don't awake the dragon...

7Iron
Aeneid
Bob
Boki
CatsEyes
Centerfield
Cheana
DirtyBird
Draco
Harkar
Houdini
JooJoo
Kathi
Melewen
Pac
Poovey
Pringle
RockClimber
SkoolGirl
Slick
Soller
SuperDuke
Tiger
Zonino

It will help if you use Brian's applet on the Argument page to test your tables before posting them!

From wHolt - 10/17/06 2:26 PM

Sunshine -
0->1 = 1
1->0 = 0

Fro- Q is still wrong
You did not use Brian's applet.
Cant make you.
So...
Do a table for A+B
Then negate the results.
Is that so hard?

Cats and Babys make us happy...

From Soller - 10/16/06 11:18 PM

A B-(A+B)=P    AB -(-A+-B)=Q
 0 0   0   0  1

 

   0 0    1    1  0
 0 1   0   1  0    0 1    1    0  0
 1 0   1   0  0    1 0    0    1  0
 1 1   1   1  0    1 1    0    0  1

                                   

 PQP -> Q   QPQ -> P
 1 0     0    0 1    1
 0 0     1    0 0    1
 0 0     1    0 0    1
 0 1     1    1 0    0

A= I have a baby.

B= He makes me happy.

P->Q -(A+B)->-(-A+-B)= If i don't have a baby nor does he makes me happy then either i do have a baby which he makes me happy.

Q->P -(-A+-B)->-(A+B)= If either i do have a baby which he makes me happy then i neither have a baby nor does he make me happy.

From Fro - 10/16/06 9:19 PM

A

B

     #1 
-A  + B

=P
  A
B 
     #7
-(A +B)
=Q
 0
 0
  1    0
 1
 
 0
 0
  1    1
 1
 0
 1
  1    1
 1
 
 0
 1
  1    0
 1
 1
 0
  0    0
 0 
 1
 0
  0    1
 1
 1
 1
  0    1
 1 
 1
 1
  0    0
 0
PQP -> Q   QPQ -> P
 1 1   1     11   1
 1 1   1     11   1
 0 1   1     10   0
 1 0   0     01   1

A: I can get this one.

B: I move on.

P→Q=(-A+B)→ -(A+B)

If I can NOT get this one or I move on THEN I can NOT get this one NOR will I move on.

Q→P= -(A+B)→(-A+B)

If I can NOT get this one NOR move on THEN I can NOT get this one or I move on.

From CatsEyes - 10/16/06 6:58 PM

 A B-(A+B)=P    AB -(A+-B)=Q
 0 0 1     1  1

 

   0 0 1      0  0
 0 1 1     0  0
    0 1 1      1  1
 1 0 0     1  0
    1 0 0       0  0
 1 1 0     0  0
    1 1 0      1  0

                                   

 PQP -> Q   QPQ -> P
 1 0     0
    0 1 1
 0 1     1
    1 0 0
 0 0     1
    0 0 1
 0 0     1
    0 0 1

A= I love my cat

B= She makes me happy

P -> Q = -(A+B) -> -(A+-B) =

If either I don't love my cat or she doesn't make me happy, then either I don't love my cat or she makes me happy

Q -> P = -(A+-B) -> -(A+B) =

If either I don't love my cat or she makes me happy, then either I don't love my cat or she does't make me happy

From Sunshine - 10/16/06 2:51 PM

#7                                                #5

A B-(A + B)=P    AB -A + -B=Q
 0 0 1      1     1

 

   0 0 1       1  1
 0 1 1      0  0    0 1 1       0  1
 1 0 0      0  0    1 0 0       1  1
 1 1 0      1  0    1 1 0       0  0

                            

 PQP -> Q   QPQ -> P
 1 1     0    1 0 0
 0 1     1    1 1 1
 0 1     1    1 0 0
 0 0     0   0  0 0

 

A: I sit in a chair

B: I sit on the floor

P-> -(A + B)

I am not sitting in a chair and I am not sitting on the floor

Q-> -A + -B

I am not sitting in a chair and I am not sitting on the floor.

From wHolt - 10/16/06 2:31 PM

Fro - pay atention to ~()
also 0+1=1, you said 0+1=0

CatsEyes - review your implication rules.

From CatsEyes - 10/16/06 9:21 AM

 A B-(A+B)=P    AB -(A+-B)=Q
 0 0 1     1  1

 

   0 0 1      0  0
 0 1 1     0  0
    0 1 1      1  1
 1 0 0     1  0
    1 0 0       0  0
 1 1 0     0  0
    1 1 0      1  0

                                   

 PQP -> Q   QPQ -> P
 1 1     0
    1 1 1
 1 1     1
    1 1 0
 1 0     1
    0 1 1
 0 1     1
    1 0 1

A= I love my cat

B= She makes me happy

P -> Q = -(A+B) -> -(A+-B) =

If either I don't love my cat or she doesn't make me happy, then either I don't love my cat or she makes me happy

Q --> P = -(A+-B) -> -(A+B) =

If either I don't love my cat or she makes me happy, then either I don't love my cat or she does't make me happy

From Fro - 10/15/06 7:53 PM

A

B

     #1 
-A  + B

=P
  A
B 
     #7
-(A +B)
=Q
 0
 0
  1    1
 1
 
 0
 0
  0    0
 0
 0
 1
  1    1
 0
 
 0
 1
  0    1
 1
 1
 0
  0    0
 1
 
 1
 0
  1    0
 1
 1
 1
  0    1
 0
 
 1
 1
  1    1
 1
PQP -> Q   QPQ -> P
 1 0   1     00   1
 0 0   1     01   0
 1 0   1     01   1
 0 1   1     01   1

A: I can get this one.

B: I move on.

P→Q=(-A+B)→ -(A+B)

If I can NOT get this one or I move on THEN I can NOT get this one NOR will I move on.

Q→P= -(A+B)→(-A+B)

If I can NOT get this one NOR move on THEN I can NOT get this one or I move on.

From JooJoo - 10/15/06 2:55 PM

Correction #3 :) 

#1 and #11

#1. -A + B #11 - (-A + -B)                                 

 A B-A + B=P    AB - (-A + -B)=Q
 0 0 1     0  1 

 

   0 0 1     1   1  0 
 0 1 1     1  1    0 1 1     0   1  0
 1 0 0     0  0    1 0 0     1   1  0
 1 1 0     1  1    1 1 0     0   0  1

 PQP -> Q   QPQ -> P
 1 0     0       0 1    1 
 1 0    0     0 1   1
 0 0    1     0 0   1
 1 1    1    1 1   1

A. Mr Smith shaved his cat

B. Mr Johnson shaved his cat

P->Q =(-A + B) -> -(-A + -B) =

If Mr Smith did not shave his cat or Mr Johnson did shave his cat, then neither Mr smith did not shave his cat or Mr Johnson did not shave his cat.

P -> Q =A Therefore P -> Q can only be true if A is true which is Mr Smith shaved his cat

Q -> P is a tautalogy therefore it can be true in all situations

- (-A + -B) -> (-A +B)

If neither Mr. Smith did not shave his cat or Mr Johnson did not shave his cat then Mr Smith did not shave his cat or Mr Johnson did shave his cat

From JooJoo - 10/15/06 1:41 PM

 

Correction #2 :) 

#1 and #11

#1. -A + B #11 - (-A + -B)                                 

 A B-A + B=P    AB - (-A + -B)=Q
 0 0 1     0  1 

 

   0 0 1           1  1 
 0 1 1     1  1    0 1 1           0    1
 1 0 0     0  0    1 0 0           1  1
 1 1 0     1  1    1 1 0           0  0

 PQP -> Q   QPQ -> P
 1 1     1       1 1    1 
 1 1    1     1 1   1
 0 1    1     1 0   0
 1 0    0    0 1   1

A. Mr Smith shaved his cat

B. Mr Johnson shaved his cat

P->Q =(-A + B) -> -(-A + -B) =

If Mr. Smith did not shave his cat or Mr Johnson shaved his cath then neither Mr Smith did not shave his cat or Mr Johnson did not shave his cat.

This P-> Q can only be true if Q is true which is  -(-A + -B)

Therefore Neither Mr Smith did not shave his cat or Mr Johnson did not shave his cat

Q-> P= -(-A +-B) -> (-A +B)

If neither Mr Smith did not shave his cat or Mr Johnson did not shave his cat, then Mr Smith did not shave his cat of Mr Johnson did shave his cat

This Q-> P can only be true if P is true which is (-A +B)

Therefore Mr Smith did not shave his cat of Mr Johnson did shave his cat.

I really hope this one is right because if it isn't then I am more lost than I thought I was!

From wHolt - 10/13/06 2:57 PM

Fro - P column not correct. 1+0=1

Aeneid - you got it!

From Aeneid - 10/12/06 8:31 PM

From Aeneid - 10/12/06 7:48 PM [Edit] [Delete]

A B-A + B=P    AB -A & B=Q
 0 0 1    0 1

 

   0 0 1    0  0
 0 1 1    1 1    0 1 1    1  1
 1 0 0    0 0    1 0 0    0  0
 1 1 0    1 1    1 1 0    1  0
 

 PQP -> Q   QPQ -> P
 10     0   01 1
 11     1   11 1
 00    1   00 1
 10    0    01 1
HELP PLEASE

A= If YOU HAVE A CLEAN CAR

B= YOU WILL GET BETTER MILAGE

P -> Q = (-A + B) -> (-A & B) =  If YOU DID NOT HAVE A CLEAN CAR OR YOU WILL GET BETTER MILAGE. THEN EITHER NOT TO HAVE A CLEAN CAR AND GET BETTER MILAGE.

Q -> P = (-A & B) -> (-A + B) = EITHER YOU DO NOT HAVE A CLEAN CAR AND GET BETTER MILAGE, THEN EITHER NOT TO HAVE A CLEAN CAR OR GET BETTER MILAGE 


Please let me know which part if any of this answer is incorrect

From Fro - 10/12/06 8:30 PM

A

B

     #1 
-A  + B

=P
  A
B 
     #7
-(A +B)
=Q
 0
 0
  1    1
 1
 
 0
 0
  0    0
 0
 0
 1
  1    1
 0
 
 0
 1
  0    1
 1
 1
 0
  0    0
 1
 
 1
 0
  1    0
 1
 1
 1
  0    1
 0
 
 1
 1
  1    1
 1
PQP -> Q   QPQ -> P
 1 0   1     00   1
 0 0   1     01   0
 1 0   0     01   1
 0 1   1     01   1

A: I can get this one.

B: I move on.

P→Q=(-A+B)→ -(A+B)

If I can NOT get this one or I move on THEN I can NOT get this one NOR will I move on.

Q→P= -(A+B)→(-A+B)

If I can NOT get this one NOR move on THEN I can NOT get this one or I move on.

From wHolt - 10/12/06 7:31 PM

TBird - rethink -a & - b
Use a 1 pixel border in your tables. Thanks. 

JooJoo - rethink your Q

From TBird - 10/11/06 8:23 PM

#6 -a & -b                   #9 -(a & b)      

 

 ab -a & - b  =q  a b -( a & b )=p 
 00  1     1 1  0 0 1     1 1
 0 1 1     0 1  01  1     0 1
 1 0 0     1 1         10  0     1 1
 1 1 0     0 0  1 1 0     0 0

 

 p→q                                     q→p 

 

 p q

 p→q 

      qp  q→p
 1 1 1  1 1 1
 1 1 1  1 1 1
 1 1 1  1 1 1
 0 0 0  0 0 0
a= i am good student

b= i make good grades

1.)  p→q   = -a & -b  → -(a & b)

if i am not a good student, or i do not make good grades, then i am not a good student, or make good grades.

2.) q→p = -(a & b)  → -a & -b

 

 

if i am not a good student, or make good grades, then i am not a good student, or i dont make good grades

From JooJoo - 10/11/06 6:44 PM

Correction :)

 

#1 and #11

#1. -A + B #11 - (-A + -B)                                 

 A B-A + B=P    AB - (-A + -B)=Q
 0 0 1     0  1 

 

   0 0 0           0   0 
 0 1 1     1  1    0 1 0           1    1
 1 0 0     0  0    1 0 1           0  1
 1 1 0     1  1    1 1 1           1  1

 PQP -> Q   QPQ -> P
 1 0     0       0 1    1 
 1 1    1     1 1   1
 0 1    1     1 0   0
 1 1    1    1 1   1

A. Mr Smith shaved his cat

B. Mr Johnson shaved his cat

P -> Q -A + B -> -(-A+-B)=

If either Mr. Smith did not shave his cat or Mr Johnson did shave his cat then nor did Mr. Smith did not shave his cat or Mr Johnson did not shave his cat

 

Q ->P -(-A +-B) -> -A + B=

If nor did Mr. Smith did not shave his cat or Mr Johnson did not shave his cat then Mr. Smith did not shave his cat or Mr Johnson did shave his cat

From wHolt - 10/10/06 11:41 AM

Reread my comment on 10/06/06
Then reread the lesson above on what => means.

From Sunshine - 10/10/06 10:43 AM

#7                                                #5

A B-(A + B)=P    AB -A + -B=Q
 0 0 1      1     1

 

   0 0 1       1  1
 0 1 1      0  0    0 1 1       0  1
 1 0 0      0  0    1 0 0       1  1
 1 1 0      1  0    1 1 0       0  0

                            

 PQP -> Q   QPQ -> P
 1 1     1    1 0 0
 0 1     0    1 1 1
 0 1     0    1 0 0
 0 0     0   0  0 0

 

A: I sit in a chair

B: I sit on the floor

P-> -(A + B)

I am not sitting in a chair and I am not sitting on the floor

Q-> -A + -B

I am not sitting in a chair and I am not sitting on the floor.

From Aeneid - 10/9/06 10:00 PM

  #1                                 #3
 A B-A + B=P    AB -A & B=Q
 0 0 1    0 1

 

   0 0 1    0  0
 0 1 1    1 1    0 1 1    1  1
 1 0 0    0 0    1 0 0    0  0
 1 1 0    1 1    1 1 0    1  0
 

 PQP -> Q   QPQ -> P
 10     1   01 1
 11     0   11 0
 00    0   00 0
 10    1    01 1
HELP PLEASE

A= If YOU HAVE A CLEAN CAR

B= YOU WILL GET BETTER MILAGE

P -> Q = (-A + B) -> (-A & B) =  If YOU DID NOT HAVE A CLEAN CAR OR YOU WILL GET BETTER MILAGE. THEN EITHER NOT TO HAVE A CLEAN CAR AND GET BETTER MILAGE.

Q -> P = (-A & B) -> (-A + B) = EITHER YOU DO NOT HAVE A CLEAN CAR AND GET BETTER MILAGE, THEN EITHER NOT TO HAVE A CLEAN CAR OR GET BETTER MILAGE 

From wHolt - 10/9/06 8:54 PM

When you resubmit, start a new comment or I will not see it or score it.

Sunshine - P->Q is still incorrect.

Soller - keep working on Q.

From Sunshine - 10/9/06 10:34 AM

ok, that was my second submission, what is wrong with it? It is coming out correct for me.

From Soller - 10/8/06 5:31 PM

A B-(A +B)=P    AB -(-A+-B)=Q
 0 0   0   0  1

 

   0 0    1   1  1
 0 1   0   1  0    0 1    1   0  1
 1 0   1   0  0    1 0    0    1  0
 1 1   1   1  0    1 1    0    0  1

                                   

 PQP -> Q   QPQ -> P
 1 1     1    1 1    1
 0 1     0    1 0   0
 0 0     1    0 0   1
 0 1     0    1 0   0

A= I have a baby.

B= He makes me happy.

P->Q -(A+B)->-(-A+-B)= If I don't have a baby nor does he make me happy then either I do have a baby which he makes me happy.

Q->P -(-A+-B)->-(A+B)= If either I do have a baby which makes me happy then I neither have a baby nor does he makes me happy.

From wHolt - 10/8/06 12:13 PM

Soller - rethink Q...

From 7Iron - 10/7/06 3:34 PM

 1.                                    3.
 A B-A  +  B=P    AB -A  &  B=Q
 0 0 1      0  1

 

   0 0 1      0  0
 0 1 1      1 1    0 1 1      1  1
 1 0 0      0  0    1 0 0      0  0
 1 1 0      1  1    1 1 0      1  0
                       

 PQP -> Q   QPQ -> P
 10   0   0 11
 1 1    1    1 1 1
 0 0    1    0 0 1
 1 0    0     0 1 1

A = I do not like animals

B = I do not like cats

P -> Q = If I like animals or I do not like cats, then I like animals and I do not like cats.

Q -> P = If I like animals and I do not like cats, then I like animals or I do not like cats.

From Soller - 10/6/06 10:07 PM

A B-(A +B)=P    AB -(-A+-B)=Q
 0 0   0   0  1

 

   0 0    1   1  1
 0 1   0   1  0    0 1    1   0  1
 1 0   1   0  0    1 0    0    1  1
 1 1   1   1  0    1 1    0    0  0

                                   

 PQP -> Q   QPQ -> P
 1 1     1    1 1    1
 0 1     0    1 0   0
 0 1     0    1 0   0
 0 0     1    0 0   1

A= I have a baby.

B= He makes me happy.

P->Q -(A+B)->-(-A+-B)= If I don't have a baby nor does he make me happy then either I do have a baby which he makes me happy.

Q->P -(-A+-B)->-(A+B)= If either I do have a baby which makes me happy then I neither have a baby nor does he makes me happy.

From wHolt - 10/6/06 1:50 PM

THIS PAGE IS NOW CLOSED TO NEWCOMERS!

x

FOR THOSE WHO STARTED, CONTINUE TILL YOU GET YOURS CORRECT.
OTHERWISE THE NEXT ASSIGNMENTS WILL BE NO FUN.
NEVERTHELESS, DO NOT WAIT TILL THE LAST DAY BEFORE MOVING ON TO THE NEXT PROBLEM.

From wHolt - 10/6/06 1:47 PM

Pod- thanks for the question.
However, 2 negatives do not always make a positive.
Does -1-7 = 17?  No.
Back in the set matcher applet you had exercises with equalities.
Review them.
Some of you forgot what you did before.
Some of your errors are not considering the ().
Some of you are not copying correctly.
Or you did not pay attention to the conditional table pattern in the lesson above.
P->Q = 1 UNLESS P=1 and Q=0.
This assignment with -> is an exercise in using symbols grammatically.
Review + and & also:
P+Q=1 UNLESS P=0 and Q=0.
P&Q=0 UNLESS P=1 and Q=1.
A ~ negates everything within the () that follow.
These simple rules are how I quickly find errors in your tables.
You can do it too.

From wHolt - 10/6/06 1:37 PM

Those below have made it past the conditional dragon —
If you are not on the list yet, go slower so you don't awake the dragon...

7Iron
Aeneid
Bob
Boki
CatsEyes
Centerfield
Cheana
DirtyBird
Draco
Harkar
Houdini
JooJoo
Kathi
Melewen
Pac
Poovey
Pringle
RockClimber
SkoolGirl
Slick
Soller
SuperDuke
Zonino

It will help if you use Brian's applet on the Argument page to test your tables before posting them!

From SkoolGirl - 10/5/06 11:18 PM

A B-A+-B=P    AB -(A+B)=Q
 0 0 1   1  1  

 

   0 0 0   0  1
 0 1 1   0  1    0 1 0   1  0
 1 0 0   1  1    1 0 1   0  0
 1 1 0   0  0    1 1 1   1  0

                                

 PQP -> Q   QPQ -> P
 1 1      1        1 1    1
 1 0     0    0 1    1
 1 0     0    0 1    1
 0 0     1    0 0    1

#5 & #7

A=Halloween is in October

B=Halloween is always on the 31st

P->Q = (-A+-B) -> -(A+B) If Halloween is not in October or Halloween is not always on the 31st then neither Halloween is in October nor Halloween is always on the 31st.

Q->P = -(A+B) -> (-A+-B) If neither Halloween is in October nor Halloween is always on the 31st then Halloween is not in October or Halloween is not always on the 31st.

From JooJoo - 10/5/06 10:29 PM

  #1 and #11

#1. -A + B #11 - (-A + -B)                                 

 A B-A + B=P    AB - (-A + -B)=Q
 0 0 1     0  1 

 

   0 0 0           0   1 
 0 1 1     1  1    0 1 0           1    1
 1 0 0     0  0    1 0 1           0  1
 1 1 0     1  1    1 1 1           1  1

 PQP -> Q   QPQ -> P
 1 1     1       1 1    1 
 1 1    1     1 1   1
 0 1    1     1 0   1
 1 1    1    1 1   1

A. Mr Smith shaved his cat

B. Mr Johnson shaved his cat

P -> Q -A & B -> -(-A+-B)=

IF Mr Smith did not shave his cat or Mr Johnson shaved his cat, then neither Mr smith did not shave his cat and Mr johnson did not shave his cat

 

Q ->P -(-A +-B) -> -A & B=

If neither Mr smith did not shave his cat and mr mr johnson did not shave his cat, then  Mr smith did not shave his cat or Mr johnson did shave his cat

From Pod - 10/5/06 9:19 PM

For example, in number 12, –(–A & –B), I understood that two negatives make a positive, which means the problem is actually A & B.  Therefore, the rest of the equations and the tables were pretty simple.  So, where did I err?

From RockClimber - 10/5/06 9:02 PM

#2) (A+ -B) & #7) -(A+B)

A B(A+ -B)=P    AB -(A+B)=Q
 0 00       11 

 

   0 00      01 
 0 10       00     0 10      10 
 1 01       11      1 01      00 
 1 11       01      1 11      1 0 

                                   

 PQP -> Q   QPQ -> P
 11 1         11 1 
 00 1         00 1 
 10 0         01 1 
 10 0         01 1 

A= I am broke.

B= I won the lottery.

P-->Q = (A+ -B) --> -(A+B)

If either I am broke or I have not won the lottery then neither am I broke nor have I won the lottery.

Q-->P = -(A+B) -->(A+ -B)

If neither I am broke nor have I won the lottery then either I am broke or I did not win the lottery.

If mine is wrong, can anyone help me please!?

From Bubba - 10/5/06 9:01 PM

ABA+-B   AB-A&B 
 0 0  0  1   1   00  1   O 0
 0 10    0      0 0  1 1   1 1
 1 0 1    1 1  1 0 0   0 0
 1 1   1  0 1  1 1 0   1 0
PQP -> Q T  QPQ -> P
 1 0      1  0 0 1 
 0 1      1  0 1 0 
 1 0      1  0 0 1 
 1 0      1  0 0 1 

NUMBERS 2&3

A = you are good at sports.

B = you will get a scholarship.

 

From DirtyBird - 10/5/06 7:55 PM

Hey TBird, I think your P->Q may be wrong down around the 0,0 part.

From TBird - 10/5/06 7:49 PM

can somone help me as`well

From DirtyBird - 10/5/06 6:44 PM

#6 & #3

 A B-A & -B=P    AB -A & B=Q
 0 0 1     1  1

 

   0 0 1     0 0
 0 1 1     0  0    0 1 1     1  1
 1 0 0     1  0    1 0 0     0  0
 1 1 0     0  0    1 1 0     1  0

 

PQP -> Q   QPQ -> P
 1 0     0      0 1    1
 0 1     1    1 0    0
 0 0     1    0 0    1
 0 0     1    0 0    1

A-I have a job.

B-I get paid.

P->Q= If I don't have a job and I don't get paid, then I don't have a job and I do get paid.

Q->P= If I don't have a job and I do get paid, then I don't have a job and I don't get paid.

From DirtyBird - 10/5/06 6:34 PM

Where is the argument applet at?

From 7Iron - 10/5/06 5:36 PM

Ok guys, I am stuck. Can someone look at mine and tell me what I am doing wrong?

7Iron

From wHolt - 10/5/06 1:17 PM

Those below have made it past the conditional dragon —
If you are not on the list yet, go slower so you don't awake the dragon...

Bob
Boki
Centerfield
Cheana
Draco
Harkar
Houdini
Kathi
Melewen
Pac
Poovey
Pringle
SuperDuke
Zonino

It will help if you use the Argument applet to test your tables before posting them!

From HotrodMinivan - 10/5/06 8:12 AM

 A B-A & B=P    AB A & -B=Q
 0 0 1     0  0

 

   0 0 0    1 0 
 0 1 1     1  1    0 1