SOLUTION: I am having a lot of trouble with word problem questions. If someone could help me that would be great. If you could show your work so that I can understand how you got the answer

Algebra ->  Conjunction -> SOLUTION: I am having a lot of trouble with word problem questions. If someone could help me that would be great. If you could show your work so that I can understand how you got the answer       Log On


   



Question 143812This question is from textbook survey of math w/applications
: I am having a lot of trouble with word problem questions. If someone could help me that would be great. If you could show your work so that I can understand how you got the answer would be apprectiated. Thanks so much!
8. (6 pts) Construct a truth table for (qp)  q
Symbols copied as squares so: 1st square is a ~ , 2nd square is /\(v reversed), 3rd square is arrow pointing to right, 4th square is ~
This question is from textbook survey of math w/applications

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
I am having a lot of trouble with word problem questions. If someone could help me that would be great. If you could show your work so that I can understand how you got the answer would be apprectiated. Thanks so much!
8. (6 pts) Construct a truth table for (qp)  q
Symbols copied as squares so: 1st square is a ~ , 2nd square is /\(v reversed), 3rd square is arrow pointing to right, 4th square is ~


We use all these

negation rules:
~T = F
~F = T

conjunction rules:
T/\T = T
T/\F = F
F/\T = F
F/\F = F

conditional rules:
T->T = T
T->F = F
F->T = T
F->F = F 

(~q/\p) -> ~q

List all the "building blocks" of that.  They are

p
q
~q
~q/\p
and end with the whole expression (~q/\p)->~q

Put these across in a table for the 4 possible cases:

        p    q    ~q    ~q/\p    (~q/\p)->~q
case 1  T    T     
case 2  T    F 
case 3  F    T
case 4  F    F

Use the negation rules and what's under q to
fill in the list under ~q
 
        p    q    ~q    ~q/\p    (~q/\p)->~q
case 1  T    T     F     
case 2  T    F     T
case 3  F    T     F
case 4  F    F     T

Use the conjunction rules and what's under ~q and under p to 
fill in the list under ~q/\p 

        p    q    ~q    ~q/\p    (~q/\p)->~q
case 1  T    T     F      F      
case 2  T    F     T      T
case 3  F    T     F      F
case 4  F    F     T      F

Finally use the conditional rules and what's under
~q/\p and under ~q to fill in the last list:

        p    q    ~q    ~q/\p    (~q/\p)->~q
case 1  T    T     F      F           T      
case 2  T    F     T      T           T 
case 3  F    T     F      F           T
case 4  F    F     T      F           T

Edwin