formal proof of $(p → q) → (¬q → ¬p)$
$begingroup$
I'm asked to give a formal proof of $(p → q) → (¬q → ¬p)$ using natural deduction. Is that like saying prove $⊢ (p → q) → (¬q → ¬p$), where it should be proved from nothing?
natural-deduction
edited Dec 29 '18 at 14:05
Key Flex
8,62261233
asked Dec 29 '18 at 13:12
Richard CameronRichard Cameron
121
$endgroup$
add a comment |
$begingroup$
I'm asked to give a formal proof of $(p → q) → (¬q → ¬p)$ using natural deduction. Is that like saying prove $⊢ (p → q) → (¬q → ¬p$), where it should be proved from nothing?
natural-deduction
edited Dec 29 '18 at 14:05
Key Flex
8,62261233
asked Dec 29 '18 at 13:12
Richard CameronRichard Cameron
121
$endgroup$
1
$begingroup$
Exactly; you have to start assuming one or more premises that you will discharge later.
$endgroup$
– Mauro ALLEGRANZA
Dec 29 '18 at 13:13
1
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Dec 29 '18 at 13:18
$begingroup$
Assume $q$ follows from $p$. Further assume not $q$. What can you say about $p$? By the way, this is called the contrapositive.
$endgroup$
– David Diaz
Dec 29 '18 at 14:11
$begingroup$
Not a duplicate, but the only answer to this question solves 95% of this problem.
$endgroup$
– Git Gud
Dec 29 '18 at 14:23
$begingroup$
What rules do you have to work with? There are many different systems of 'natural deduction', each with their own set of rules.
$endgroup$
– Bram28
Dec 29 '18 at 18:13
add a comment |
$begingroup$
I'm asked to give a formal proof of $(p → q) → (¬q → ¬p)$ using natural deduction. Is that like saying prove $⊢ (p → q) → (¬q → ¬p$), where it should be proved from nothing?
natural-deduction
edited Dec 29 '18 at 14:05
Key Flex
8,62261233
asked Dec 29 '18 at 13:12
Richard CameronRichard Cameron
121
$endgroup$
I'm asked to give a formal proof of $(p → q) → (¬q → ¬p)$ using natural deduction. Is that like saying prove $⊢ (p → q) → (¬q → ¬p$), where it should be proved from nothing?
natural-deduction
natural-deduction
edited Dec 29 '18 at 14:05
Key Flex
8,62261233
asked Dec 29 '18 at 13:12
Richard CameronRichard Cameron
121
edited Dec 29 '18 at 14:05
Key Flex
8,62261233
asked Dec 29 '18 at 13:12
Richard CameronRichard Cameron
121
edited Dec 29 '18 at 14:05
Key Flex
8,62261233
edited Dec 29 '18 at 14:05
Key Flex
8,62261233
edited Dec 29 '18 at 14:05
Key Flex
8,62261233
8,62261233
asked Dec 29 '18 at 13:12
Richard CameronRichard Cameron
121
asked Dec 29 '18 at 13:12
Richard CameronRichard Cameron
121
asked Dec 29 '18 at 13:12
Richard CameronRichard Cameron
121
121
1
$begingroup$
Exactly; you have to start assuming one or more premises that you will discharge later.
$endgroup$
– Mauro ALLEGRANZA
Dec 29 '18 at 13:13
1
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Dec 29 '18 at 13:18
$begingroup$
Assume $q$ follows from $p$. Further assume not $q$. What can you say about $p$? By the way, this is called the contrapositive.
$endgroup$
– David Diaz
Dec 29 '18 at 14:11
$begingroup$
Not a duplicate, but the only answer to this question solves 95% of this problem.
$endgroup$
– Git Gud
Dec 29 '18 at 14:23
$begingroup$
What rules do you have to work with? There are many different systems of 'natural deduction', each with their own set of rules.
$endgroup$
– Bram28
Dec 29 '18 at 18:13
add a comment |
1
$begingroup$
Exactly; you have to start assuming one or more premises that you will discharge later.
$endgroup$
– Mauro ALLEGRANZA
Dec 29 '18 at 13:13
1
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Dec 29 '18 at 13:18
$begingroup$
Assume $q$ follows from $p$. Further assume not $q$. What can you say about $p$? By the way, this is called the contrapositive.
$endgroup$
– David Diaz
Dec 29 '18 at 14:11
$begingroup$
Not a duplicate, but the only answer to this question solves 95% of this problem.
$endgroup$
– Git Gud
Dec 29 '18 at 14:23
$begingroup$
What rules do you have to work with? There are many different systems of 'natural deduction', each with their own set of rules.
$endgroup$
– Bram28
Dec 29 '18 at 18:13
1
1
$begingroup$
Exactly; you have to start assuming one or more premises that you will discharge later.
$endgroup$
– Mauro ALLEGRANZA
Dec 29 '18 at 13:13
$begingroup$
Exactly; you have to start assuming one or more premises that you will discharge later.
$endgroup$
– Mauro ALLEGRANZA
Dec 29 '18 at 13:13
1
1
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Dec 29 '18 at 13:18
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Dec 29 '18 at 13:18
$begingroup$
Assume $q$ follows from $p$. Further assume not $q$. What can you say about $p$? By the way, this is called the contrapositive.
$endgroup$
– David Diaz
Dec 29 '18 at 14:11
$begingroup$
Assume $q$ follows from $p$. Further assume not $q$. What can you say about $p$? By the way, this is called the contrapositive.
$endgroup$
– David Diaz
Dec 29 '18 at 14:11
$begingroup$
Not a duplicate, but the only answer to this question solves 95% of this problem.
$endgroup$
– Git Gud
Dec 29 '18 at 14:23
$begingroup$
Not a duplicate, but the only answer to this question solves 95% of this problem.
$endgroup$
– Git Gud
Dec 29 '18 at 14:23
$begingroup$
What rules do you have to work with? There are many different systems of 'natural deduction', each with their own set of rules.
$endgroup$
– Bram28
Dec 29 '18 at 18:13
$begingroup$
What rules do you have to work with? There are many different systems of 'natural deduction', each with their own set of rules.
$endgroup$
– Bram28
Dec 29 '18 at 18:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: the following statements are all equivalent:
- $pto q$
- $neg plor q$
- $negneg qlorneg p$
- $neg qtoneg p$
answered Dec 29 '18 at 13:33
J.G.J.G.
30.9k23149
$endgroup$
1
$begingroup$
The formal proof that comes naturally to me uses none of this. How is this helpful?
$endgroup$
– Git Gud
Dec 29 '18 at 14:20
$begingroup$
@GitGud If you have a different route, well done. But to any future reader considering this problem who doesn't, the use comes in providing a sequence of statements worth proving equivalent. How you stitch them together into a proof depends on your preferred notation.
$endgroup$
– J.G.
Dec 29 '18 at 15:59
$begingroup$
I didn't say what I should have said. How do you go from this to a formal proof? As I see it, you need to go to the moon and back, I don't see how this can be helpful.
$endgroup$
– Git Gud
Dec 29 '18 at 16:09
add a comment |
$begingroup$
The following proof uses modus tollens (MT):
However, one can derive the modus tollens rule in the following way. This uses the proof provided on page 138 of forallx linked to below along with a link to the proof checker used here:
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
answered Jan 21 at 0:43
Frank HubenyFrank Hubeny
5042519
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: the following statements are all equivalent:
- $pto q$
- $neg plor q$
- $negneg qlorneg p$
- $neg qtoneg p$
answered Dec 29 '18 at 13:33
J.G.J.G.
30.9k23149
$endgroup$
1
$begingroup$
The formal proof that comes naturally to me uses none of this. How is this helpful?
$endgroup$
– Git Gud
Dec 29 '18 at 14:20
$begingroup$
@GitGud If you have a different route, well done. But to any future reader considering this problem who doesn't, the use comes in providing a sequence of statements worth proving equivalent. How you stitch them together into a proof depends on your preferred notation.
$endgroup$
– J.G.
Dec 29 '18 at 15:59
$begingroup$
I didn't say what I should have said. How do you go from this to a formal proof? As I see it, you need to go to the moon and back, I don't see how this can be helpful.
$endgroup$
– Git Gud
Dec 29 '18 at 16:09
add a comment |
$begingroup$
Hint: the following statements are all equivalent:
- $pto q$
- $neg plor q$
- $negneg qlorneg p$
- $neg qtoneg p$
answered Dec 29 '18 at 13:33
J.G.J.G.
30.9k23149
$endgroup$
1
$begingroup$
The formal proof that comes naturally to me uses none of this. How is this helpful?
$endgroup$
– Git Gud
Dec 29 '18 at 14:20
$begingroup$
@GitGud If you have a different route, well done. But to any future reader considering this problem who doesn't, the use comes in providing a sequence of statements worth proving equivalent. How you stitch them together into a proof depends on your preferred notation.
$endgroup$
– J.G.
Dec 29 '18 at 15:59
$begingroup$
I didn't say what I should have said. How do you go from this to a formal proof? As I see it, you need to go to the moon and back, I don't see how this can be helpful.
$endgroup$
– Git Gud
Dec 29 '18 at 16:09
add a comment |
$begingroup$
Hint: the following statements are all equivalent:
- $pto q$
- $neg plor q$
- $negneg qlorneg p$
- $neg qtoneg p$
answered Dec 29 '18 at 13:33
J.G.J.G.
30.9k23149
$endgroup$
Hint: the following statements are all equivalent:
- $pto q$
- $neg plor q$
- $negneg qlorneg p$
- $neg qtoneg p$
answered Dec 29 '18 at 13:33
J.G.J.G.
30.9k23149
answered Dec 29 '18 at 13:33
J.G.J.G.
30.9k23149
answered Dec 29 '18 at 13:33
J.G.J.G.
30.9k23149
answered Dec 29 '18 at 13:33
J.G.J.G.
30.9k23149
30.9k23149
1
$begingroup$
The formal proof that comes naturally to me uses none of this. How is this helpful?
$endgroup$
– Git Gud
Dec 29 '18 at 14:20
$begingroup$
@GitGud If you have a different route, well done. But to any future reader considering this problem who doesn't, the use comes in providing a sequence of statements worth proving equivalent. How you stitch them together into a proof depends on your preferred notation.
$endgroup$
– J.G.
Dec 29 '18 at 15:59
$begingroup$
I didn't say what I should have said. How do you go from this to a formal proof? As I see it, you need to go to the moon and back, I don't see how this can be helpful.
$endgroup$
– Git Gud
Dec 29 '18 at 16:09
add a comment |
1
$begingroup$
The formal proof that comes naturally to me uses none of this. How is this helpful?
$endgroup$
– Git Gud
Dec 29 '18 at 14:20
$begingroup$
@GitGud If you have a different route, well done. But to any future reader considering this problem who doesn't, the use comes in providing a sequence of statements worth proving equivalent. How you stitch them together into a proof depends on your preferred notation.
$endgroup$
– J.G.
Dec 29 '18 at 15:59
$begingroup$
I didn't say what I should have said. How do you go from this to a formal proof? As I see it, you need to go to the moon and back, I don't see how this can be helpful.
$endgroup$
– Git Gud
Dec 29 '18 at 16:09
1
1
$begingroup$
The formal proof that comes naturally to me uses none of this. How is this helpful?
$endgroup$
– Git Gud
Dec 29 '18 at 14:20
$begingroup$
The formal proof that comes naturally to me uses none of this. How is this helpful?
$endgroup$
– Git Gud
Dec 29 '18 at 14:20
$begingroup$
@GitGud If you have a different route, well done. But to any future reader considering this problem who doesn't, the use comes in providing a sequence of statements worth proving equivalent. How you stitch them together into a proof depends on your preferred notation.
$endgroup$
– J.G.
Dec 29 '18 at 15:59
$begingroup$
@GitGud If you have a different route, well done. But to any future reader considering this problem who doesn't, the use comes in providing a sequence of statements worth proving equivalent. How you stitch them together into a proof depends on your preferred notation.
$endgroup$
– J.G.
Dec 29 '18 at 15:59
$begingroup$
I didn't say what I should have said. How do you go from this to a formal proof? As I see it, you need to go to the moon and back, I don't see how this can be helpful.
$endgroup$
– Git Gud
Dec 29 '18 at 16:09
$begingroup$
I didn't say what I should have said. How do you go from this to a formal proof? As I see it, you need to go to the moon and back, I don't see how this can be helpful.
$endgroup$
– Git Gud
Dec 29 '18 at 16:09
add a comment |
$begingroup$
The following proof uses modus tollens (MT):
However, one can derive the modus tollens rule in the following way. This uses the proof provided on page 138 of forallx linked to below along with a link to the proof checker used here:
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
answered Jan 21 at 0:43
Frank HubenyFrank Hubeny
5042519
$endgroup$
add a comment |
$begingroup$
The following proof uses modus tollens (MT):
However, one can derive the modus tollens rule in the following way. This uses the proof provided on page 138 of forallx linked to below along with a link to the proof checker used here:
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
answered Jan 21 at 0:43
Frank HubenyFrank Hubeny
5042519
$endgroup$
add a comment |
$begingroup$
The following proof uses modus tollens (MT):
However, one can derive the modus tollens rule in the following way. This uses the proof provided on page 138 of forallx linked to below along with a link to the proof checker used here:
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
answered Jan 21 at 0:43
Frank HubenyFrank Hubeny
5042519
$endgroup$
The following proof uses modus tollens (MT):
However, one can derive the modus tollens rule in the following way. This uses the proof provided on page 138 of forallx linked to below along with a link to the proof checker used here:
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
answered Jan 21 at 0:43
Frank HubenyFrank Hubeny
5042519
answered Jan 21 at 0:43
Frank HubenyFrank Hubeny
5042519
answered Jan 21 at 0:43
Frank HubenyFrank Hubeny
5042519
answered Jan 21 at 0:43
Frank HubenyFrank Hubeny
5042519
5042519
add a comment |
add a comment |
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1
$begingroup$
Exactly; you have to start assuming one or more premises that you will discharge later.
$endgroup$
– Mauro ALLEGRANZA
Dec 29 '18 at 13:13
1
$begingroup$
Please use MathJax in future.
$endgroup$
– Shaun
Dec 29 '18 at 13:18
$begingroup$
Assume $q$ follows from $p$. Further assume not $q$. What can you say about $p$? By the way, this is called the contrapositive.
$endgroup$
– David Diaz
Dec 29 '18 at 14:11
$begingroup$
Not a duplicate, but the only answer to this question solves 95% of this problem.
$endgroup$
– Git Gud
Dec 29 '18 at 14:23
$begingroup$
What rules do you have to work with? There are many different systems of 'natural deduction', each with their own set of rules.
$endgroup$
– Bram28
Dec 29 '18 at 18:13