why you should follow it? [duplicate]
This question already has an answer here:
In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?
21 answers
We learned expression of deduce, i.e. =>.
But now I dont have capable reason for I agree I represent True if assumption is False.
Anybady there having to explain reason for its deduce?
Best regards,
logic
marked as duplicate by Shaun, gammatester, Hans Lundmark, spaceisdarkgreen, Derek Elkins Nov 24 at 22:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 1 more comment
This question already has an answer here:
In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?
21 answers
We learned expression of deduce, i.e. =>.
But now I dont have capable reason for I agree I represent True if assumption is False.
Anybady there having to explain reason for its deduce?
Best regards,
logic
marked as duplicate by Shaun, gammatester, Hans Lundmark, spaceisdarkgreen, Derek Elkins Nov 24 at 22:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Do you mean to ask why $P implies Q$ is true when $P$ is false?
– quid♦
Nov 24 at 20:44
1
Yes, of course.
– quickybrown
Nov 24 at 20:47
2
@quickybrown Then you should have written that - what you've written right now isn't clear at all.
– Noah Schweber
Nov 24 at 20:51
In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
– quickybrown
Nov 24 at 21:02
@quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
– spaceisdarkgreen
Nov 24 at 22:04
|
show 1 more comment
This question already has an answer here:
In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?
21 answers
We learned expression of deduce, i.e. =>.
But now I dont have capable reason for I agree I represent True if assumption is False.
Anybady there having to explain reason for its deduce?
Best regards,
logic
This question already has an answer here:
In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?
21 answers
We learned expression of deduce, i.e. =>.
But now I dont have capable reason for I agree I represent True if assumption is False.
Anybady there having to explain reason for its deduce?
Best regards,
This question already has an answer here:
In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?
21 answers
logic
logic
edited Nov 24 at 21:22
asked Nov 24 at 20:35
quickybrown
13
13
marked as duplicate by Shaun, gammatester, Hans Lundmark, spaceisdarkgreen, Derek Elkins Nov 24 at 22:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Shaun, gammatester, Hans Lundmark, spaceisdarkgreen, Derek Elkins Nov 24 at 22:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Do you mean to ask why $P implies Q$ is true when $P$ is false?
– quid♦
Nov 24 at 20:44
1
Yes, of course.
– quickybrown
Nov 24 at 20:47
2
@quickybrown Then you should have written that - what you've written right now isn't clear at all.
– Noah Schweber
Nov 24 at 20:51
In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
– quickybrown
Nov 24 at 21:02
@quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
– spaceisdarkgreen
Nov 24 at 22:04
|
show 1 more comment
2
Do you mean to ask why $P implies Q$ is true when $P$ is false?
– quid♦
Nov 24 at 20:44
1
Yes, of course.
– quickybrown
Nov 24 at 20:47
2
@quickybrown Then you should have written that - what you've written right now isn't clear at all.
– Noah Schweber
Nov 24 at 20:51
In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
– quickybrown
Nov 24 at 21:02
@quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
– spaceisdarkgreen
Nov 24 at 22:04
2
2
Do you mean to ask why $P implies Q$ is true when $P$ is false?
– quid♦
Nov 24 at 20:44
Do you mean to ask why $P implies Q$ is true when $P$ is false?
– quid♦
Nov 24 at 20:44
1
1
Yes, of course.
– quickybrown
Nov 24 at 20:47
Yes, of course.
– quickybrown
Nov 24 at 20:47
2
2
@quickybrown Then you should have written that - what you've written right now isn't clear at all.
– Noah Schweber
Nov 24 at 20:51
@quickybrown Then you should have written that - what you've written right now isn't clear at all.
– Noah Schweber
Nov 24 at 20:51
In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
– quickybrown
Nov 24 at 21:02
In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
– quickybrown
Nov 24 at 21:02
@quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
– spaceisdarkgreen
Nov 24 at 22:04
@quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
– spaceisdarkgreen
Nov 24 at 22:04
|
show 1 more comment
2 Answers
2
active
oldest
votes
Here is a table that should help
begin{array}{|c|c|c|}
hline
P & Q & PRightarrow Q \ hline
T & T & T \ hline
T & F & F \ hline
F & T & T \ hline
F & F & T \ hline
end{array}
For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.
As an extension to this exercise try this one to solidify your understanding:
begin{array}{|c|c|c|}
hline
P & Q & QRightarrow P \ hline
T & T & \ hline
T & F & \ hline
F & T & \ hline
F & F & \ hline
end{array}
add a comment |
One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.
But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
– quickybrown
Nov 24 at 21:17
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here is a table that should help
begin{array}{|c|c|c|}
hline
P & Q & PRightarrow Q \ hline
T & T & T \ hline
T & F & F \ hline
F & T & T \ hline
F & F & T \ hline
end{array}
For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.
As an extension to this exercise try this one to solidify your understanding:
begin{array}{|c|c|c|}
hline
P & Q & QRightarrow P \ hline
T & T & \ hline
T & F & \ hline
F & T & \ hline
F & F & \ hline
end{array}
add a comment |
Here is a table that should help
begin{array}{|c|c|c|}
hline
P & Q & PRightarrow Q \ hline
T & T & T \ hline
T & F & F \ hline
F & T & T \ hline
F & F & T \ hline
end{array}
For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.
As an extension to this exercise try this one to solidify your understanding:
begin{array}{|c|c|c|}
hline
P & Q & QRightarrow P \ hline
T & T & \ hline
T & F & \ hline
F & T & \ hline
F & F & \ hline
end{array}
add a comment |
Here is a table that should help
begin{array}{|c|c|c|}
hline
P & Q & PRightarrow Q \ hline
T & T & T \ hline
T & F & F \ hline
F & T & T \ hline
F & F & T \ hline
end{array}
For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.
As an extension to this exercise try this one to solidify your understanding:
begin{array}{|c|c|c|}
hline
P & Q & QRightarrow P \ hline
T & T & \ hline
T & F & \ hline
F & T & \ hline
F & F & \ hline
end{array}
Here is a table that should help
begin{array}{|c|c|c|}
hline
P & Q & PRightarrow Q \ hline
T & T & T \ hline
T & F & F \ hline
F & T & T \ hline
F & F & T \ hline
end{array}
For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.
As an extension to this exercise try this one to solidify your understanding:
begin{array}{|c|c|c|}
hline
P & Q & QRightarrow P \ hline
T & T & \ hline
T & F & \ hline
F & T & \ hline
F & F & \ hline
end{array}
answered Nov 24 at 21:01
Wolfy
2,28011038
2,28011038
add a comment |
add a comment |
One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.
But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
– quickybrown
Nov 24 at 21:17
add a comment |
One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.
But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
– quickybrown
Nov 24 at 21:17
add a comment |
One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.
One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.
answered Nov 24 at 21:09
Nodt Greenish
29813
29813
But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
– quickybrown
Nov 24 at 21:17
add a comment |
But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
– quickybrown
Nov 24 at 21:17
But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
– quickybrown
Nov 24 at 21:17
But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
– quickybrown
Nov 24 at 21:17
add a comment |
2
Do you mean to ask why $P implies Q$ is true when $P$ is false?
– quid♦
Nov 24 at 20:44
1
Yes, of course.
– quickybrown
Nov 24 at 20:47
2
@quickybrown Then you should have written that - what you've written right now isn't clear at all.
– Noah Schweber
Nov 24 at 20:51
In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
– quickybrown
Nov 24 at 21:02
@quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
– spaceisdarkgreen
Nov 24 at 22:04