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How does the value matrix correspond to the truth table of an operation?
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I have been self studying mathematical logic through A Concise Introduction to Mathematical Logic by Wolfgang Rautenberg and got stuck. I am unable to understand how the value matrix corresponds with the truth table of a given operation for instance the AND operation. I know that in a matrix the place an element is placed into important. Is there some kind of relation based on the position in the matrix. For example the ith column and jth row. Why are the values ordered as they are within the matrix? 
matrices logic proof-explanation education boolean-algebra
asked Aug 6 at 6:39
user372382
726
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up vote
2
down vote
favorite
I have been self studying mathematical logic through A Concise Introduction to Mathematical Logic by Wolfgang Rautenberg and got stuck. I am unable to understand how the value matrix corresponds with the truth table of a given operation for instance the AND operation. I know that in a matrix the place an element is placed into important. Is there some kind of relation based on the position in the matrix. For example the ith column and jth row. Why are the values ordered as they are within the matrix?
matrices logic proof-explanation education boolean-algebra
asked Aug 6 at 6:39
user372382
726
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have been self studying mathematical logic through A Concise Introduction to Mathematical Logic by Wolfgang Rautenberg and got stuck. I am unable to understand how the value matrix corresponds with the truth table of a given operation for instance the AND operation. I know that in a matrix the place an element is placed into important. Is there some kind of relation based on the position in the matrix. For example the ith column and jth row. Why are the values ordered as they are within the matrix?
matrices logic proof-explanation education boolean-algebra
asked Aug 6 at 6:39
user372382
726
I have been self studying mathematical logic through A Concise Introduction to Mathematical Logic by Wolfgang Rautenberg and got stuck. I am unable to understand how the value matrix corresponds with the truth table of a given operation for instance the AND operation. I know that in a matrix the place an element is placed into important. Is there some kind of relation based on the position in the matrix. For example the ith column and jth row. Why are the values ordered as they are within the matrix?
matrices logic proof-explanation education boolean-algebra
asked Aug 6 at 6:39
user372382
726
asked Aug 6 at 6:39
user372382
726
asked Aug 6 at 6:39
user372382
726
asked Aug 6 at 6:39
user372382
726
726
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
The author is using an abbreviated form of the notation often used for binary operations on finite algebraic structures, leaving out the row and column labels. Drawn up in full, the table for conjunction would look like this:
$$
beginarraycc
land & 1 & 0 \hline
1 & 1 & 0 \
0 & 0 & 1
endarray
$$
The idea is that to evaluate $a land b$, you look for the row labelled $a$ and the column labelled $b$ and the result is the corresponding entry in the matrix: $1 land 1 = 1$, $1 land 0 = 0$, $0 land 1 = 0$ and $0 land 0 = 0$.
answered Aug 6 at 10:49
Rob Arthan
27.1k42863
Thank you for your answer, it makes more sense.In essence the matrix is shorthand for a Cayley table while defining the operation on the set 0,1. My concern though is that in a Cayley table the order of the rows and columns does not matter you could have also written 0 1 on the top of the table as well as easily as 1 0 and still have a valid operation. However for a matrix the position of the element within the matrix is very important. Putting 0 in position 1,1 results in a different matrix from putting a 1 in position 1,1. How is the positions of the elements determined within the matrix?
– user372382
Aug 7 at 6:54
@user372382: yes, you're right: the author's notation is so abbreviated that it is ambiguous. It's standard to list the elements in the same order in the row and column headings, but you have to use your knowledge of what conjunction etc. "ought to mean" to figure out the order the author actually has in mind (unless he has written something about his conventions for what he calls a truth table).
– Rob Arthan
Aug 7 at 17:38
add a comment |Â
up vote
0
down vote
Looks like the rows,1,2 go with the truth values, T,F of p, and columns 1,2 go with the truth values T,F of q in the connective p (connecvtive) q.
Thus row 1 column 2 corresponds to p true (T) and q false (F), etc.
answered Aug 6 at 7:08
coffeemath
1,2711213
1
You seem to be right. I never used it for 2 variable system, but I think it can be seen as a 2 variable Karnaugh map.
– F.Carette
Aug 6 at 9:28
@F.Carette Only difference is in the linked system they put $0,1$ ((F,T) for the rows and similarly for the columns.
– coffeemath
Aug 6 at 17:15
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The author is using an abbreviated form of the notation often used for binary operations on finite algebraic structures, leaving out the row and column labels. Drawn up in full, the table for conjunction would look like this:
$$
beginarraycc
land & 1 & 0 \hline
1 & 1 & 0 \
0 & 0 & 1
endarray
$$
The idea is that to evaluate $a land b$, you look for the row labelled $a$ and the column labelled $b$ and the result is the corresponding entry in the matrix: $1 land 1 = 1$, $1 land 0 = 0$, $0 land 1 = 0$ and $0 land 0 = 0$.
answered Aug 6 at 10:49
Rob Arthan
27.1k42863
Thank you for your answer, it makes more sense.In essence the matrix is shorthand for a Cayley table while defining the operation on the set 0,1. My concern though is that in a Cayley table the order of the rows and columns does not matter you could have also written 0 1 on the top of the table as well as easily as 1 0 and still have a valid operation. However for a matrix the position of the element within the matrix is very important. Putting 0 in position 1,1 results in a different matrix from putting a 1 in position 1,1. How is the positions of the elements determined within the matrix?
– user372382
Aug 7 at 6:54
@user372382: yes, you're right: the author's notation is so abbreviated that it is ambiguous. It's standard to list the elements in the same order in the row and column headings, but you have to use your knowledge of what conjunction etc. "ought to mean" to figure out the order the author actually has in mind (unless he has written something about his conventions for what he calls a truth table).
– Rob Arthan
Aug 7 at 17:38
add a comment |Â
up vote
1
down vote
The author is using an abbreviated form of the notation often used for binary operations on finite algebraic structures, leaving out the row and column labels. Drawn up in full, the table for conjunction would look like this:
$$
beginarraycc
land & 1 & 0 \hline
1 & 1 & 0 \
0 & 0 & 1
endarray
$$
The idea is that to evaluate $a land b$, you look for the row labelled $a$ and the column labelled $b$ and the result is the corresponding entry in the matrix: $1 land 1 = 1$, $1 land 0 = 0$, $0 land 1 = 0$ and $0 land 0 = 0$.
answered Aug 6 at 10:49
Rob Arthan
27.1k42863
Thank you for your answer, it makes more sense.In essence the matrix is shorthand for a Cayley table while defining the operation on the set 0,1. My concern though is that in a Cayley table the order of the rows and columns does not matter you could have also written 0 1 on the top of the table as well as easily as 1 0 and still have a valid operation. However for a matrix the position of the element within the matrix is very important. Putting 0 in position 1,1 results in a different matrix from putting a 1 in position 1,1. How is the positions of the elements determined within the matrix?
– user372382
Aug 7 at 6:54
@user372382: yes, you're right: the author's notation is so abbreviated that it is ambiguous. It's standard to list the elements in the same order in the row and column headings, but you have to use your knowledge of what conjunction etc. "ought to mean" to figure out the order the author actually has in mind (unless he has written something about his conventions for what he calls a truth table).
– Rob Arthan
Aug 7 at 17:38
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The author is using an abbreviated form of the notation often used for binary operations on finite algebraic structures, leaving out the row and column labels. Drawn up in full, the table for conjunction would look like this:
$$
beginarraycc
land & 1 & 0 \hline
1 & 1 & 0 \
0 & 0 & 1
endarray
$$
The idea is that to evaluate $a land b$, you look for the row labelled $a$ and the column labelled $b$ and the result is the corresponding entry in the matrix: $1 land 1 = 1$, $1 land 0 = 0$, $0 land 1 = 0$ and $0 land 0 = 0$.
answered Aug 6 at 10:49
Rob Arthan
27.1k42863
The author is using an abbreviated form of the notation often used for binary operations on finite algebraic structures, leaving out the row and column labels. Drawn up in full, the table for conjunction would look like this:
$$
beginarraycc
land & 1 & 0 \hline
1 & 1 & 0 \
0 & 0 & 1
endarray
$$
The idea is that to evaluate $a land b$, you look for the row labelled $a$ and the column labelled $b$ and the result is the corresponding entry in the matrix: $1 land 1 = 1$, $1 land 0 = 0$, $0 land 1 = 0$ and $0 land 0 = 0$.
answered Aug 6 at 10:49
Rob Arthan
27.1k42863
answered Aug 6 at 10:49
Rob Arthan
27.1k42863
answered Aug 6 at 10:49
Rob Arthan
27.1k42863
answered Aug 6 at 10:49
Rob Arthan
27.1k42863
27.1k42863
Thank you for your answer, it makes more sense.In essence the matrix is shorthand for a Cayley table while defining the operation on the set 0,1. My concern though is that in a Cayley table the order of the rows and columns does not matter you could have also written 0 1 on the top of the table as well as easily as 1 0 and still have a valid operation. However for a matrix the position of the element within the matrix is very important. Putting 0 in position 1,1 results in a different matrix from putting a 1 in position 1,1. How is the positions of the elements determined within the matrix?
– user372382
Aug 7 at 6:54
@user372382: yes, you're right: the author's notation is so abbreviated that it is ambiguous. It's standard to list the elements in the same order in the row and column headings, but you have to use your knowledge of what conjunction etc. "ought to mean" to figure out the order the author actually has in mind (unless he has written something about his conventions for what he calls a truth table).
– Rob Arthan
Aug 7 at 17:38
add a comment |Â
Thank you for your answer, it makes more sense.In essence the matrix is shorthand for a Cayley table while defining the operation on the set 0,1. My concern though is that in a Cayley table the order of the rows and columns does not matter you could have also written 0 1 on the top of the table as well as easily as 1 0 and still have a valid operation. However for a matrix the position of the element within the matrix is very important. Putting 0 in position 1,1 results in a different matrix from putting a 1 in position 1,1. How is the positions of the elements determined within the matrix?
– user372382
Aug 7 at 6:54
@user372382: yes, you're right: the author's notation is so abbreviated that it is ambiguous. It's standard to list the elements in the same order in the row and column headings, but you have to use your knowledge of what conjunction etc. "ought to mean" to figure out the order the author actually has in mind (unless he has written something about his conventions for what he calls a truth table).
– Rob Arthan
Aug 7 at 17:38
Thank you for your answer, it makes more sense.In essence the matrix is shorthand for a Cayley table while defining the operation on the set 0,1. My concern though is that in a Cayley table the order of the rows and columns does not matter you could have also written 0 1 on the top of the table as well as easily as 1 0 and still have a valid operation. However for a matrix the position of the element within the matrix is very important. Putting 0 in position 1,1 results in a different matrix from putting a 1 in position 1,1. How is the positions of the elements determined within the matrix?
– user372382
Aug 7 at 6:54
Thank you for your answer, it makes more sense.In essence the matrix is shorthand for a Cayley table while defining the operation on the set 0,1. My concern though is that in a Cayley table the order of the rows and columns does not matter you could have also written 0 1 on the top of the table as well as easily as 1 0 and still have a valid operation. However for a matrix the position of the element within the matrix is very important. Putting 0 in position 1,1 results in a different matrix from putting a 1 in position 1,1. How is the positions of the elements determined within the matrix?
– user372382
Aug 7 at 6:54
@user372382: yes, you're right: the author's notation is so abbreviated that it is ambiguous. It's standard to list the elements in the same order in the row and column headings, but you have to use your knowledge of what conjunction etc. "ought to mean" to figure out the order the author actually has in mind (unless he has written something about his conventions for what he calls a truth table).
– Rob Arthan
Aug 7 at 17:38
@user372382: yes, you're right: the author's notation is so abbreviated that it is ambiguous. It's standard to list the elements in the same order in the row and column headings, but you have to use your knowledge of what conjunction etc. "ought to mean" to figure out the order the author actually has in mind (unless he has written something about his conventions for what he calls a truth table).
– Rob Arthan
Aug 7 at 17:38
add a comment |Â
up vote
0
down vote
Looks like the rows,1,2 go with the truth values, T,F of p, and columns 1,2 go with the truth values T,F of q in the connective p (connecvtive) q.
Thus row 1 column 2 corresponds to p true (T) and q false (F), etc.
answered Aug 6 at 7:08
coffeemath
1,2711213
1
You seem to be right. I never used it for 2 variable system, but I think it can be seen as a 2 variable Karnaugh map.
– F.Carette
Aug 6 at 9:28
@F.Carette Only difference is in the linked system they put $0,1$ ((F,T) for the rows and similarly for the columns.
– coffeemath
Aug 6 at 17:15
add a comment |Â
up vote
0
down vote
Looks like the rows,1,2 go with the truth values, T,F of p, and columns 1,2 go with the truth values T,F of q in the connective p (connecvtive) q.
Thus row 1 column 2 corresponds to p true (T) and q false (F), etc.
answered Aug 6 at 7:08
coffeemath
1,2711213
1
You seem to be right. I never used it for 2 variable system, but I think it can be seen as a 2 variable Karnaugh map.
– F.Carette
Aug 6 at 9:28
@F.Carette Only difference is in the linked system they put $0,1$ ((F,T) for the rows and similarly for the columns.
– coffeemath
Aug 6 at 17:15
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Looks like the rows,1,2 go with the truth values, T,F of p, and columns 1,2 go with the truth values T,F of q in the connective p (connecvtive) q.
Thus row 1 column 2 corresponds to p true (T) and q false (F), etc.
answered Aug 6 at 7:08
coffeemath
1,2711213
Looks like the rows,1,2 go with the truth values, T,F of p, and columns 1,2 go with the truth values T,F of q in the connective p (connecvtive) q.
Thus row 1 column 2 corresponds to p true (T) and q false (F), etc.
answered Aug 6 at 7:08
coffeemath
1,2711213
edited Aug 6 at 7:18
answered Aug 6 at 7:08
coffeemath
1,2711213
answered Aug 6 at 7:08
coffeemath
1,2711213
answered Aug 6 at 7:08
coffeemath
1,2711213
1,2711213
1
You seem to be right. I never used it for 2 variable system, but I think it can be seen as a 2 variable Karnaugh map.
– F.Carette
Aug 6 at 9:28
@F.Carette Only difference is in the linked system they put $0,1$ ((F,T) for the rows and similarly for the columns.
– coffeemath
Aug 6 at 17:15
add a comment |Â
1
You seem to be right. I never used it for 2 variable system, but I think it can be seen as a 2 variable Karnaugh map.
– F.Carette
Aug 6 at 9:28
@F.Carette Only difference is in the linked system they put $0,1$ ((F,T) for the rows and similarly for the columns.
– coffeemath
Aug 6 at 17:15
1
1
You seem to be right. I never used it for 2 variable system, but I think it can be seen as a 2 variable Karnaugh map.
– F.Carette
Aug 6 at 9:28
You seem to be right. I never used it for 2 variable system, but I think it can be seen as a 2 variable Karnaugh map.
– F.Carette
Aug 6 at 9:28
@F.Carette Only difference is in the linked system they put $0,1$ ((F,T) for the rows and similarly for the columns.
– coffeemath
Aug 6 at 17:15
@F.Carette Only difference is in the linked system they put $0,1$ ((F,T) for the rows and similarly for the columns.
– coffeemath
Aug 6 at 17:15
add a comment |Â
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