Discrete Mathematics - Truth Function Problem
I keep trying to reason how is letter C the answer to this question, but I can't fully understand how can one find the solution to this type of problems. What I understand:
There's 2 arguments in the given truth function, therefore we have 2^(2) = 4 total statements. This is because we have f(x, y) = k. In the case we had 3 arguments/variables, then we would have f(x, y, z) = k.
The f( T, F) = T is the only statement that results true for when the values T, F are assigned to the respective variables. (i.e. In the case P = T and Q = F, or Q = T, and P = F).
For any compound statement the function F(T, F) has to result in true.
The algorithm for TRUE function:
Input: A truth function F with an n number of arguments x_1, x_2, x_3, ..., x_n.
Output: A statement S whose truth function is f. This statement can only involve the connectives of conjunction "AND", disjunction "OR" , and negation "~".
Step 1:
Consider a basic statement q_i for each argument x_i of f, where i = 1,2,3, .., n.
Step 2:
For each combination of values for the arguments x_1, x_2, x_3, ..., x_n. Consider the statement:
S_j = P^(1)_j "AND" P^(2)_j "AND"... "AND" P^(n)_j.
where, P^i _ j {
q_i IF x_i = T
~q_i IF x_i = F
}
Step 3:
Let J_1, J_2, J_3, ..., J_k. Be the index for the combination of values for x_1, x_2, x_3, ..., x_n that force f to produce T.
S = S_ J_1 "OR" S_ J_2 "OR" S_ J_3
CLAIM I: Every statement S_J is only true for the combination of values that we used to construct it.
CLAIM II: For each combination of values, there is only one S_J that is true for that combination.
PICTURE OF THE PROBLEM HERE:
https://imgur.com/a/jnMckeM
Sorry for the notation, I don't know any text editors for this sort of syntax. I hope is readable.
Thank you in advance!
discrete-mathematics logic propositional-calculus
asked Nov 19 '18 at 0:40
ChalupaBatmac
1
add a comment |
I keep trying to reason how is letter C the answer to this question, but I can't fully understand how can one find the solution to this type of problems. What I understand:
There's 2 arguments in the given truth function, therefore we have 2^(2) = 4 total statements. This is because we have f(x, y) = k. In the case we had 3 arguments/variables, then we would have f(x, y, z) = k.
The f( T, F) = T is the only statement that results true for when the values T, F are assigned to the respective variables. (i.e. In the case P = T and Q = F, or Q = T, and P = F).
For any compound statement the function F(T, F) has to result in true.
The algorithm for TRUE function:
Input: A truth function F with an n number of arguments x_1, x_2, x_3, ..., x_n.
Output: A statement S whose truth function is f. This statement can only involve the connectives of conjunction "AND", disjunction "OR" , and negation "~".
Step 1:
Consider a basic statement q_i for each argument x_i of f, where i = 1,2,3, .., n.
Step 2:
For each combination of values for the arguments x_1, x_2, x_3, ..., x_n. Consider the statement:
S_j = P^(1)_j "AND" P^(2)_j "AND"... "AND" P^(n)_j.
where, P^i _ j {
q_i IF x_i = T
~q_i IF x_i = F
}
Step 3:
Let J_1, J_2, J_3, ..., J_k. Be the index for the combination of values for x_1, x_2, x_3, ..., x_n that force f to produce T.
S = S_ J_1 "OR" S_ J_2 "OR" S_ J_3
CLAIM I: Every statement S_J is only true for the combination of values that we used to construct it.
CLAIM II: For each combination of values, there is only one S_J that is true for that combination.
PICTURE OF THE PROBLEM HERE:
https://imgur.com/a/jnMckeM
Sorry for the notation, I don't know any text editors for this sort of syntax. I hope is readable.
Thank you in advance!
discrete-mathematics logic propositional-calculus
asked Nov 19 '18 at 0:40
ChalupaBatmac
1
add a comment |
I keep trying to reason how is letter C the answer to this question, but I can't fully understand how can one find the solution to this type of problems. What I understand:
There's 2 arguments in the given truth function, therefore we have 2^(2) = 4 total statements. This is because we have f(x, y) = k. In the case we had 3 arguments/variables, then we would have f(x, y, z) = k.
The f( T, F) = T is the only statement that results true for when the values T, F are assigned to the respective variables. (i.e. In the case P = T and Q = F, or Q = T, and P = F).
For any compound statement the function F(T, F) has to result in true.
The algorithm for TRUE function:
Input: A truth function F with an n number of arguments x_1, x_2, x_3, ..., x_n.
Output: A statement S whose truth function is f. This statement can only involve the connectives of conjunction "AND", disjunction "OR" , and negation "~".
Step 1:
Consider a basic statement q_i for each argument x_i of f, where i = 1,2,3, .., n.
Step 2:
For each combination of values for the arguments x_1, x_2, x_3, ..., x_n. Consider the statement:
S_j = P^(1)_j "AND" P^(2)_j "AND"... "AND" P^(n)_j.
where, P^i _ j {
q_i IF x_i = T
~q_i IF x_i = F
}
Step 3:
Let J_1, J_2, J_3, ..., J_k. Be the index for the combination of values for x_1, x_2, x_3, ..., x_n that force f to produce T.
S = S_ J_1 "OR" S_ J_2 "OR" S_ J_3
CLAIM I: Every statement S_J is only true for the combination of values that we used to construct it.
CLAIM II: For each combination of values, there is only one S_J that is true for that combination.
PICTURE OF THE PROBLEM HERE:
https://imgur.com/a/jnMckeM
Sorry for the notation, I don't know any text editors for this sort of syntax. I hope is readable.
Thank you in advance!
discrete-mathematics logic propositional-calculus
asked Nov 19 '18 at 0:40
ChalupaBatmac
1
I keep trying to reason how is letter C the answer to this question, but I can't fully understand how can one find the solution to this type of problems. What I understand:
There's 2 arguments in the given truth function, therefore we have 2^(2) = 4 total statements. This is because we have f(x, y) = k. In the case we had 3 arguments/variables, then we would have f(x, y, z) = k.
The f( T, F) = T is the only statement that results true for when the values T, F are assigned to the respective variables. (i.e. In the case P = T and Q = F, or Q = T, and P = F).
For any compound statement the function F(T, F) has to result in true.
The algorithm for TRUE function:
Input: A truth function F with an n number of arguments x_1, x_2, x_3, ..., x_n.
Output: A statement S whose truth function is f. This statement can only involve the connectives of conjunction "AND", disjunction "OR" , and negation "~".
Step 1:
Consider a basic statement q_i for each argument x_i of f, where i = 1,2,3, .., n.
Step 2:
For each combination of values for the arguments x_1, x_2, x_3, ..., x_n. Consider the statement:
S_j = P^(1)_j "AND" P^(2)_j "AND"... "AND" P^(n)_j.
where, P^i _ j {
q_i IF x_i = T
~q_i IF x_i = F
}
Step 3:
Let J_1, J_2, J_3, ..., J_k. Be the index for the combination of values for x_1, x_2, x_3, ..., x_n that force f to produce T.
S = S_ J_1 "OR" S_ J_2 "OR" S_ J_3
CLAIM I: Every statement S_J is only true for the combination of values that we used to construct it.
CLAIM II: For each combination of values, there is only one S_J that is true for that combination.
PICTURE OF THE PROBLEM HERE:
https://imgur.com/a/jnMckeM
Sorry for the notation, I don't know any text editors for this sort of syntax. I hope is readable.
Thank you in advance!
discrete-mathematics logic propositional-calculus
discrete-mathematics logic propositional-calculus
asked Nov 19 '18 at 0:40
ChalupaBatmac
1
asked Nov 19 '18 at 0:40
ChalupaBatmac
1
asked Nov 19 '18 at 0:40
ChalupaBatmac
1
asked Nov 19 '18 at 0:40
ChalupaBatmac
1
asked Nov 19 '18 at 0:40
ChalupaBatmac
1
1
add a comment |
add a comment |
1 Answer
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The problem uses $f(p,q)$, so $p$ is the first argument, and $q$ the second. So, the truth table shows that the only case where $f$ returns True is if $p=T$ and $q=F$. If $q=T$ and $p=F$ then $f(p,q)=F$. So, what you have at the very end of the second bullet point is wrong. (And maybe now it is easier to see what the correct answer is!)
answered Nov 19 '18 at 1:37
Bram28
60.2k44590
Do we assume p corresponds to the first argument, x_1, or is this based on the results of the truth table formed?
– ChalupaBatmac
Nov 19 '18 at 16:25
@ChalupaBatmac Yes, it is assumed that $p$ refers to the first argument, and $q$ to the second.
– Bram28
Nov 19 '18 at 16:27
Ok, so then it comes down to verifying the truth values of each compound statement using the values given by the function f. Thank you for the quick response.
– ChalupaBatmac
Nov 19 '18 at 16:29
@ChalupaBatmac Yes ... should be easy. Glad I could help!
– Bram28
Nov 19 '18 at 17:06
add a comment |
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The problem uses $f(p,q)$, so $p$ is the first argument, and $q$ the second. So, the truth table shows that the only case where $f$ returns True is if $p=T$ and $q=F$. If $q=T$ and $p=F$ then $f(p,q)=F$. So, what you have at the very end of the second bullet point is wrong. (And maybe now it is easier to see what the correct answer is!)
answered Nov 19 '18 at 1:37
Bram28
60.2k44590
Do we assume p corresponds to the first argument, x_1, or is this based on the results of the truth table formed?
– ChalupaBatmac
Nov 19 '18 at 16:25
@ChalupaBatmac Yes, it is assumed that $p$ refers to the first argument, and $q$ to the second.
– Bram28
Nov 19 '18 at 16:27
Ok, so then it comes down to verifying the truth values of each compound statement using the values given by the function f. Thank you for the quick response.
– ChalupaBatmac
Nov 19 '18 at 16:29
@ChalupaBatmac Yes ... should be easy. Glad I could help!
– Bram28
Nov 19 '18 at 17:06
add a comment |
The problem uses $f(p,q)$, so $p$ is the first argument, and $q$ the second. So, the truth table shows that the only case where $f$ returns True is if $p=T$ and $q=F$. If $q=T$ and $p=F$ then $f(p,q)=F$. So, what you have at the very end of the second bullet point is wrong. (And maybe now it is easier to see what the correct answer is!)
answered Nov 19 '18 at 1:37
Bram28
60.2k44590
Do we assume p corresponds to the first argument, x_1, or is this based on the results of the truth table formed?
– ChalupaBatmac
Nov 19 '18 at 16:25
@ChalupaBatmac Yes, it is assumed that $p$ refers to the first argument, and $q$ to the second.
– Bram28
Nov 19 '18 at 16:27
Ok, so then it comes down to verifying the truth values of each compound statement using the values given by the function f. Thank you for the quick response.
– ChalupaBatmac
Nov 19 '18 at 16:29
@ChalupaBatmac Yes ... should be easy. Glad I could help!
– Bram28
Nov 19 '18 at 17:06
add a comment |
The problem uses $f(p,q)$, so $p$ is the first argument, and $q$ the second. So, the truth table shows that the only case where $f$ returns True is if $p=T$ and $q=F$. If $q=T$ and $p=F$ then $f(p,q)=F$. So, what you have at the very end of the second bullet point is wrong. (And maybe now it is easier to see what the correct answer is!)
answered Nov 19 '18 at 1:37
Bram28
60.2k44590
The problem uses $f(p,q)$, so $p$ is the first argument, and $q$ the second. So, the truth table shows that the only case where $f$ returns True is if $p=T$ and $q=F$. If $q=T$ and $p=F$ then $f(p,q)=F$. So, what you have at the very end of the second bullet point is wrong. (And maybe now it is easier to see what the correct answer is!)
answered Nov 19 '18 at 1:37
Bram28
60.2k44590
answered Nov 19 '18 at 1:37
Bram28
60.2k44590
answered Nov 19 '18 at 1:37
Bram28
60.2k44590
answered Nov 19 '18 at 1:37
Bram28
60.2k44590
60.2k44590
Do we assume p corresponds to the first argument, x_1, or is this based on the results of the truth table formed?
– ChalupaBatmac
Nov 19 '18 at 16:25
@ChalupaBatmac Yes, it is assumed that $p$ refers to the first argument, and $q$ to the second.
– Bram28
Nov 19 '18 at 16:27
Ok, so then it comes down to verifying the truth values of each compound statement using the values given by the function f. Thank you for the quick response.
– ChalupaBatmac
Nov 19 '18 at 16:29
@ChalupaBatmac Yes ... should be easy. Glad I could help!
– Bram28
Nov 19 '18 at 17:06
add a comment |
Do we assume p corresponds to the first argument, x_1, or is this based on the results of the truth table formed?
– ChalupaBatmac
Nov 19 '18 at 16:25
@ChalupaBatmac Yes, it is assumed that $p$ refers to the first argument, and $q$ to the second.
– Bram28
Nov 19 '18 at 16:27
Ok, so then it comes down to verifying the truth values of each compound statement using the values given by the function f. Thank you for the quick response.
– ChalupaBatmac
Nov 19 '18 at 16:29
@ChalupaBatmac Yes ... should be easy. Glad I could help!
– Bram28
Nov 19 '18 at 17:06
Do we assume p corresponds to the first argument, x_1, or is this based on the results of the truth table formed?
– ChalupaBatmac
Nov 19 '18 at 16:25
Do we assume p corresponds to the first argument, x_1, or is this based on the results of the truth table formed?
– ChalupaBatmac
Nov 19 '18 at 16:25
@ChalupaBatmac Yes, it is assumed that $p$ refers to the first argument, and $q$ to the second.
– Bram28
Nov 19 '18 at 16:27
@ChalupaBatmac Yes, it is assumed that $p$ refers to the first argument, and $q$ to the second.
– Bram28
Nov 19 '18 at 16:27
Ok, so then it comes down to verifying the truth values of each compound statement using the values given by the function f. Thank you for the quick response.
– ChalupaBatmac
Nov 19 '18 at 16:29
Ok, so then it comes down to verifying the truth values of each compound statement using the values given by the function f. Thank you for the quick response.
– ChalupaBatmac
Nov 19 '18 at 16:29
@ChalupaBatmac Yes ... should be easy. Glad I could help!
– Bram28
Nov 19 '18 at 17:06
@ChalupaBatmac Yes ... should be easy. Glad I could help!
– Bram28
Nov 19 '18 at 17:06
add a comment |
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