Mixing
The assertion : A implies non(A), is it false?
$begingroup$
I have proved that $(1) ; A to B.$
But I want prove that:
$(2); B to neg A.$
From (2), is (2) a false assertion? Because if (2) is true, this means that:
$A to neg A = lnot A lor lnot A equiv lnot A$ is true.
logic propositional-calculus
edited Jan 25 at 16:42
Namaste
1
asked Jan 25 at 12:00
Martin LutherMartin Luther
11
$endgroup$
add a comment |
$begingroup$
I have proved that $(1) ; A to B.$
But I want prove that:
$(2); B to neg A.$
From (2), is (2) a false assertion? Because if (2) is true, this means that:
$A to neg A = lnot A lor lnot A equiv lnot A$ is true.
logic propositional-calculus
edited Jan 25 at 16:42
Namaste
1
asked Jan 25 at 12:00
Martin LutherMartin Luther
11
$endgroup$
$begingroup$
If (1) and (2) are correct then (3) is also correct by transitivity of implication, which would suggest A is false
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
If $A$ is false then $A to B$, $Bto lnot A$ and $A to lnot A$ will all be true. But $A to lnot A$ can only be true if $A$ is false. If you have successfully proven $A to B$, I do not know if you can successfully prove $B to lnot A$, but if you can you will have successfully proven $lnot A$.
$endgroup$
– fleablood
Jan 25 at 16:55
add a comment |
$begingroup$
I have proved that $(1) ; A to B.$
But I want prove that:
$(2); B to neg A.$
From (2), is (2) a false assertion? Because if (2) is true, this means that:
$A to neg A = lnot A lor lnot A equiv lnot A$ is true.
logic propositional-calculus
edited Jan 25 at 16:42
Namaste
1
asked Jan 25 at 12:00
Martin LutherMartin Luther
11
$endgroup$
I have proved that $(1) ; A to B.$
But I want prove that:
$(2); B to neg A.$
From (2), is (2) a false assertion? Because if (2) is true, this means that:
$A to neg A = lnot A lor lnot A equiv lnot A$ is true.
logic propositional-calculus
logic propositional-calculus
edited Jan 25 at 16:42
Namaste
1
asked Jan 25 at 12:00
Martin LutherMartin Luther
11
edited Jan 25 at 16:42
Namaste
1
asked Jan 25 at 12:00
Martin LutherMartin Luther
11
edited Jan 25 at 16:42
Namaste
1
edited Jan 25 at 16:42
Namaste
1
edited Jan 25 at 16:42
Namaste
1
1
asked Jan 25 at 12:00
Martin LutherMartin Luther
11
asked Jan 25 at 12:00
Martin LutherMartin Luther
11
asked Jan 25 at 12:00
Martin LutherMartin Luther
11
11
$begingroup$
If (1) and (2) are correct then (3) is also correct by transitivity of implication, which would suggest A is false
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
If $A$ is false then $A to B$, $Bto lnot A$ and $A to lnot A$ will all be true. But $A to lnot A$ can only be true if $A$ is false. If you have successfully proven $A to B$, I do not know if you can successfully prove $B to lnot A$, but if you can you will have successfully proven $lnot A$.
$endgroup$
– fleablood
Jan 25 at 16:55
add a comment |
$begingroup$
If (1) and (2) are correct then (3) is also correct by transitivity of implication, which would suggest A is false
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
If $A$ is false then $A to B$, $Bto lnot A$ and $A to lnot A$ will all be true. But $A to lnot A$ can only be true if $A$ is false. If you have successfully proven $A to B$, I do not know if you can successfully prove $B to lnot A$, but if you can you will have successfully proven $lnot A$.
$endgroup$
– fleablood
Jan 25 at 16:55
$begingroup$
If (1) and (2) are correct then (3) is also correct by transitivity of implication, which would suggest A is false
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
If (1) and (2) are correct then (3) is also correct by transitivity of implication, which would suggest A is false
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
If $A$ is false then $A to B$, $Bto lnot A$ and $A to lnot A$ will all be true. But $A to lnot A$ can only be true if $A$ is false. If you have successfully proven $A to B$, I do not know if you can successfully prove $B to lnot A$, but if you can you will have successfully proven $lnot A$.
$endgroup$
– fleablood
Jan 25 at 16:55
$begingroup$
If $A$ is false then $A to B$, $Bto lnot A$ and $A to lnot A$ will all be true. But $A to lnot A$ can only be true if $A$ is false. If you have successfully proven $A to B$, I do not know if you can successfully prove $B to lnot A$, but if you can you will have successfully proven $lnot A$.
$endgroup$
– fleablood
Jan 25 at 16:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
So we start with the proven implication:
begin{align*}
&A Rightarrow B \
end{align*}
We then consider the new implications:
begin{align*}
B Rightarrow !A\
A Rightarrow !A \
end{align*}
This proves that A has to be False, because False statements can imply True statements, but True statements cannot imply False statements.
answered Jan 25 at 12:06
GustavGustav
1469
$endgroup$
$begingroup$
False statements can imply true statements, so if A is false then there is no fallacy
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
@Henry That is true, but the same statement can only be true or false, and here A has to be both true and false.
$endgroup$
– Gustav
Jan 25 at 12:07
$begingroup$
No. You can't conclude $B$ from $Ato B$ if $A$ is false. So $Ato lnot A$ isn't a fallacy, it's a proof that $A$ is false.
$endgroup$
– MJD
Jan 25 at 12:18
$begingroup$
@MJD That is true, but he does not say that he has B. He says that B implies not A.
$endgroup$
– Gustav
Jan 25 at 12:25
$begingroup$
@MJD You are right. I thought I knew how this all worked, guess not. Do you want to add this as an answer, or should I add it to my answer?
$endgroup$
– Gustav
Jan 25 at 12:30
|
show 6 more comments
$begingroup$
From (2), is (2) a false assertion? Because if (2) is true, this means that:
A→¬A=¬A∨¬A≡¬A
is true.
Why would proving $lnot A$ be a contradiction? Were you told that $A$ had to be true?
....
If not, then proving $A to B$ and $B to lnot A$ will prove that $A$ is false. Which isn't a contradiction.
Notice $P to Q$ and $P to lnot Q$ is not a contradiction. The can both be true but they are both true if and only if $P$ is false.
Likewise $P to lnot P$ is also not a contradiction. It will be true if and only $P$ is false.
So $A to B$ and $B to lnot A$ will have proven $A to lnot A$ which isn't a contradiction if $A$ is false. However if you prove that both $A to B$ and $B to lnot A$, that that is a proof that $A$ is false
The following may be useful:
$P to lnot P iff lnot P$
$P to Q iff lnot Q to lnot P$
$(Pto Q)land (Pto lnot Q) iff lnot Q$
$(P to Q) land (lnot P to Q)iff Q$
$(P to Q) land (Q to lnot P) iff lnot P$.
etc.
Also if $Q$ is true then $X to Q$ is always true.
And if $P$ is false then $P to X$ is always true.
So for any $X$ then $A to X$ and $Xto lnot A$ will be consistent and will both be true whenever $A$ is false (but if $A$ is true then one will be true and the other false depending upon whether $X$ is true or false.)
answered Jan 25 at 17:30
fleabloodfleablood
72.8k22788
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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oldest
votes
$begingroup$
So we start with the proven implication:
begin{align*}
&A Rightarrow B \
end{align*}
We then consider the new implications:
begin{align*}
B Rightarrow !A\
A Rightarrow !A \
end{align*}
This proves that A has to be False, because False statements can imply True statements, but True statements cannot imply False statements.
answered Jan 25 at 12:06
GustavGustav
1469
$endgroup$
$begingroup$
False statements can imply true statements, so if A is false then there is no fallacy
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
@Henry That is true, but the same statement can only be true or false, and here A has to be both true and false.
$endgroup$
– Gustav
Jan 25 at 12:07
$begingroup$
No. You can't conclude $B$ from $Ato B$ if $A$ is false. So $Ato lnot A$ isn't a fallacy, it's a proof that $A$ is false.
$endgroup$
– MJD
Jan 25 at 12:18
$begingroup$
@MJD That is true, but he does not say that he has B. He says that B implies not A.
$endgroup$
– Gustav
Jan 25 at 12:25
$begingroup$
@MJD You are right. I thought I knew how this all worked, guess not. Do you want to add this as an answer, or should I add it to my answer?
$endgroup$
– Gustav
Jan 25 at 12:30
|
show 6 more comments
$begingroup$
So we start with the proven implication:
begin{align*}
&A Rightarrow B \
end{align*}
We then consider the new implications:
begin{align*}
B Rightarrow !A\
A Rightarrow !A \
end{align*}
This proves that A has to be False, because False statements can imply True statements, but True statements cannot imply False statements.
answered Jan 25 at 12:06
GustavGustav
1469
$endgroup$
$begingroup$
False statements can imply true statements, so if A is false then there is no fallacy
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
@Henry That is true, but the same statement can only be true or false, and here A has to be both true and false.
$endgroup$
– Gustav
Jan 25 at 12:07
$begingroup$
No. You can't conclude $B$ from $Ato B$ if $A$ is false. So $Ato lnot A$ isn't a fallacy, it's a proof that $A$ is false.
$endgroup$
– MJD
Jan 25 at 12:18
$begingroup$
@MJD That is true, but he does not say that he has B. He says that B implies not A.
$endgroup$
– Gustav
Jan 25 at 12:25
$begingroup$
@MJD You are right. I thought I knew how this all worked, guess not. Do you want to add this as an answer, or should I add it to my answer?
$endgroup$
– Gustav
Jan 25 at 12:30
|
show 6 more comments
$begingroup$
So we start with the proven implication:
begin{align*}
&A Rightarrow B \
end{align*}
We then consider the new implications:
begin{align*}
B Rightarrow !A\
A Rightarrow !A \
end{align*}
This proves that A has to be False, because False statements can imply True statements, but True statements cannot imply False statements.
answered Jan 25 at 12:06
GustavGustav
1469
$endgroup$
So we start with the proven implication:
begin{align*}
&A Rightarrow B \
end{align*}
We then consider the new implications:
begin{align*}
B Rightarrow !A\
A Rightarrow !A \
end{align*}
This proves that A has to be False, because False statements can imply True statements, but True statements cannot imply False statements.
answered Jan 25 at 12:06
GustavGustav
1469
edited Jan 25 at 12:38
answered Jan 25 at 12:06
GustavGustav
1469
answered Jan 25 at 12:06
GustavGustav
1469
answered Jan 25 at 12:06
GustavGustav
1469
1469
$begingroup$
False statements can imply true statements, so if A is false then there is no fallacy
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
@Henry That is true, but the same statement can only be true or false, and here A has to be both true and false.
$endgroup$
– Gustav
Jan 25 at 12:07
$begingroup$
No. You can't conclude $B$ from $Ato B$ if $A$ is false. So $Ato lnot A$ isn't a fallacy, it's a proof that $A$ is false.
$endgroup$
– MJD
Jan 25 at 12:18
$begingroup$
@MJD That is true, but he does not say that he has B. He says that B implies not A.
$endgroup$
– Gustav
Jan 25 at 12:25
$begingroup$
@MJD You are right. I thought I knew how this all worked, guess not. Do you want to add this as an answer, or should I add it to my answer?
$endgroup$
– Gustav
Jan 25 at 12:30
|
show 6 more comments
$begingroup$
False statements can imply true statements, so if A is false then there is no fallacy
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
@Henry That is true, but the same statement can only be true or false, and here A has to be both true and false.
$endgroup$
– Gustav
Jan 25 at 12:07
$begingroup$
No. You can't conclude $B$ from $Ato B$ if $A$ is false. So $Ato lnot A$ isn't a fallacy, it's a proof that $A$ is false.
$endgroup$
– MJD
Jan 25 at 12:18
$begingroup$
@MJD That is true, but he does not say that he has B. He says that B implies not A.
$endgroup$
– Gustav
Jan 25 at 12:25
$begingroup$
@MJD You are right. I thought I knew how this all worked, guess not. Do you want to add this as an answer, or should I add it to my answer?
$endgroup$
– Gustav
Jan 25 at 12:30
$begingroup$
False statements can imply true statements, so if A is false then there is no fallacy
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
False statements can imply true statements, so if A is false then there is no fallacy
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
@Henry That is true, but the same statement can only be true or false, and here A has to be both true and false.
$endgroup$
– Gustav
Jan 25 at 12:07
$begingroup$
@Henry That is true, but the same statement can only be true or false, and here A has to be both true and false.
$endgroup$
– Gustav
Jan 25 at 12:07
$begingroup$
No. You can't conclude $B$ from $Ato B$ if $A$ is false. So $Ato lnot A$ isn't a fallacy, it's a proof that $A$ is false.
$endgroup$
– MJD
Jan 25 at 12:18
$begingroup$
No. You can't conclude $B$ from $Ato B$ if $A$ is false. So $Ato lnot A$ isn't a fallacy, it's a proof that $A$ is false.
$endgroup$
– MJD
Jan 25 at 12:18
$begingroup$
@MJD That is true, but he does not say that he has B. He says that B implies not A.
$endgroup$
– Gustav
Jan 25 at 12:25
$begingroup$
@MJD That is true, but he does not say that he has B. He says that B implies not A.
$endgroup$
– Gustav
Jan 25 at 12:25
$begingroup$
@MJD You are right. I thought I knew how this all worked, guess not. Do you want to add this as an answer, or should I add it to my answer?
$endgroup$
– Gustav
Jan 25 at 12:30
$begingroup$
@MJD You are right. I thought I knew how this all worked, guess not. Do you want to add this as an answer, or should I add it to my answer?
$endgroup$
– Gustav
Jan 25 at 12:30
|
show 6 more comments
$begingroup$
From (2), is (2) a false assertion? Because if (2) is true, this means that:
A→¬A=¬A∨¬A≡¬A
is true.
Why would proving $lnot A$ be a contradiction? Were you told that $A$ had to be true?
....
If not, then proving $A to B$ and $B to lnot A$ will prove that $A$ is false. Which isn't a contradiction.
Notice $P to Q$ and $P to lnot Q$ is not a contradiction. The can both be true but they are both true if and only if $P$ is false.
Likewise $P to lnot P$ is also not a contradiction. It will be true if and only $P$ is false.
So $A to B$ and $B to lnot A$ will have proven $A to lnot A$ which isn't a contradiction if $A$ is false. However if you prove that both $A to B$ and $B to lnot A$, that that is a proof that $A$ is false
The following may be useful:
$P to lnot P iff lnot P$
$P to Q iff lnot Q to lnot P$
$(Pto Q)land (Pto lnot Q) iff lnot Q$
$(P to Q) land (lnot P to Q)iff Q$
$(P to Q) land (Q to lnot P) iff lnot P$.
etc.
Also if $Q$ is true then $X to Q$ is always true.
And if $P$ is false then $P to X$ is always true.
So for any $X$ then $A to X$ and $Xto lnot A$ will be consistent and will both be true whenever $A$ is false (but if $A$ is true then one will be true and the other false depending upon whether $X$ is true or false.)
answered Jan 25 at 17:30
fleabloodfleablood
72.8k22788
$endgroup$
add a comment |
$begingroup$
From (2), is (2) a false assertion? Because if (2) is true, this means that:
A→¬A=¬A∨¬A≡¬A
is true.
Why would proving $lnot A$ be a contradiction? Were you told that $A$ had to be true?
....
If not, then proving $A to B$ and $B to lnot A$ will prove that $A$ is false. Which isn't a contradiction.
Notice $P to Q$ and $P to lnot Q$ is not a contradiction. The can both be true but they are both true if and only if $P$ is false.
Likewise $P to lnot P$ is also not a contradiction. It will be true if and only $P$ is false.
So $A to B$ and $B to lnot A$ will have proven $A to lnot A$ which isn't a contradiction if $A$ is false. However if you prove that both $A to B$ and $B to lnot A$, that that is a proof that $A$ is false
The following may be useful:
$P to lnot P iff lnot P$
$P to Q iff lnot Q to lnot P$
$(Pto Q)land (Pto lnot Q) iff lnot Q$
$(P to Q) land (lnot P to Q)iff Q$
$(P to Q) land (Q to lnot P) iff lnot P$.
etc.
Also if $Q$ is true then $X to Q$ is always true.
And if $P$ is false then $P to X$ is always true.
So for any $X$ then $A to X$ and $Xto lnot A$ will be consistent and will both be true whenever $A$ is false (but if $A$ is true then one will be true and the other false depending upon whether $X$ is true or false.)
answered Jan 25 at 17:30
fleabloodfleablood
72.8k22788
$endgroup$
add a comment |
$begingroup$
From (2), is (2) a false assertion? Because if (2) is true, this means that:
A→¬A=¬A∨¬A≡¬A
is true.
Why would proving $lnot A$ be a contradiction? Were you told that $A$ had to be true?
....
If not, then proving $A to B$ and $B to lnot A$ will prove that $A$ is false. Which isn't a contradiction.
Notice $P to Q$ and $P to lnot Q$ is not a contradiction. The can both be true but they are both true if and only if $P$ is false.
Likewise $P to lnot P$ is also not a contradiction. It will be true if and only $P$ is false.
So $A to B$ and $B to lnot A$ will have proven $A to lnot A$ which isn't a contradiction if $A$ is false. However if you prove that both $A to B$ and $B to lnot A$, that that is a proof that $A$ is false
The following may be useful:
$P to lnot P iff lnot P$
$P to Q iff lnot Q to lnot P$
$(Pto Q)land (Pto lnot Q) iff lnot Q$
$(P to Q) land (lnot P to Q)iff Q$
$(P to Q) land (Q to lnot P) iff lnot P$.
etc.
Also if $Q$ is true then $X to Q$ is always true.
And if $P$ is false then $P to X$ is always true.
So for any $X$ then $A to X$ and $Xto lnot A$ will be consistent and will both be true whenever $A$ is false (but if $A$ is true then one will be true and the other false depending upon whether $X$ is true or false.)
answered Jan 25 at 17:30
fleabloodfleablood
72.8k22788
$endgroup$
From (2), is (2) a false assertion? Because if (2) is true, this means that:
A→¬A=¬A∨¬A≡¬A
is true.
Why would proving $lnot A$ be a contradiction? Were you told that $A$ had to be true?
....
If not, then proving $A to B$ and $B to lnot A$ will prove that $A$ is false. Which isn't a contradiction.
Notice $P to Q$ and $P to lnot Q$ is not a contradiction. The can both be true but they are both true if and only if $P$ is false.
Likewise $P to lnot P$ is also not a contradiction. It will be true if and only $P$ is false.
So $A to B$ and $B to lnot A$ will have proven $A to lnot A$ which isn't a contradiction if $A$ is false. However if you prove that both $A to B$ and $B to lnot A$, that that is a proof that $A$ is false
The following may be useful:
$P to lnot P iff lnot P$
$P to Q iff lnot Q to lnot P$
$(Pto Q)land (Pto lnot Q) iff lnot Q$
$(P to Q) land (lnot P to Q)iff Q$
$(P to Q) land (Q to lnot P) iff lnot P$.
etc.
Also if $Q$ is true then $X to Q$ is always true.
And if $P$ is false then $P to X$ is always true.
So for any $X$ then $A to X$ and $Xto lnot A$ will be consistent and will both be true whenever $A$ is false (but if $A$ is true then one will be true and the other false depending upon whether $X$ is true or false.)
answered Jan 25 at 17:30
fleabloodfleablood
72.8k22788
answered Jan 25 at 17:30
fleabloodfleablood
72.8k22788
answered Jan 25 at 17:30
fleabloodfleablood
72.8k22788
answered Jan 25 at 17:30
fleabloodfleablood
72.8k22788
72.8k22788
add a comment |
add a comment |
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$begingroup$
If (1) and (2) are correct then (3) is also correct by transitivity of implication, which would suggest A is false
$endgroup$
– Henry
Jan 25 at 12:06
$begingroup$
If $A$ is false then $A to B$, $Bto lnot A$ and $A to lnot A$ will all be true. But $A to lnot A$ can only be true if $A$ is false. If you have successfully proven $A to B$, I do not know if you can successfully prove $B to lnot A$, but if you can you will have successfully proven $lnot A$.
$endgroup$
– fleablood
Jan 25 at 16:55