How to prove ‘∃xP(x)’ from ‘¬∀x(P(x)→Q(x))’
up vote
3
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What would a formal Fitch proof for this look like?
I am given ¬∀x(P(x)→Q(x)), and need to derive ∃xP(x) from it.
I started with this, but I don't know if I am doing the right thing, and where to go from there:
EDIT: Did it (see answer)
logic proof fitch quantification
asked Nov 25 at 19:56
35308
827
add a comment |
up vote
3
down vote
favorite
What would a formal Fitch proof for this look like?
I am given ¬∀x(P(x)→Q(x)), and need to derive ∃xP(x) from it.
I started with this, but I don't know if I am doing the right thing, and where to go from there:
EDIT: Did it (see answer)
logic proof fitch quantification
asked Nov 25 at 19:56
35308
827
So far no answers written out in English. Isn't the crux of the matter that if there were no $x$ such that $P(x)$, then $P(x)$ would be false for all $x$ and thus would imply absolutely anything?
– Wildcard
Nov 26 at 5:33
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
What would a formal Fitch proof for this look like?
I am given ¬∀x(P(x)→Q(x)), and need to derive ∃xP(x) from it.
I started with this, but I don't know if I am doing the right thing, and where to go from there:
EDIT: Did it (see answer)
logic proof fitch quantification
asked Nov 25 at 19:56
35308
827
What would a formal Fitch proof for this look like?
I am given ¬∀x(P(x)→Q(x)), and need to derive ∃xP(x) from it.
I started with this, but I don't know if I am doing the right thing, and where to go from there:
EDIT: Did it (see answer)
logic proof fitch quantification
logic proof fitch quantification
asked Nov 25 at 19:56
35308
827
asked Nov 25 at 19:56
35308
827
edited Nov 25 at 20:58
asked Nov 25 at 19:56
35308
827
asked Nov 25 at 19:56
35308
827
asked Nov 25 at 19:56
35308
827
827
So far no answers written out in English. Isn't the crux of the matter that if there were no $x$ such that $P(x)$, then $P(x)$ would be false for all $x$ and thus would imply absolutely anything?
– Wildcard
Nov 26 at 5:33
add a comment |
So far no answers written out in English. Isn't the crux of the matter that if there were no $x$ such that $P(x)$, then $P(x)$ would be false for all $x$ and thus would imply absolutely anything?
– Wildcard
Nov 26 at 5:33
So far no answers written out in English. Isn't the crux of the matter that if there were no $x$ such that $P(x)$, then $P(x)$ would be false for all $x$ and thus would imply absolutely anything?
– Wildcard
Nov 26 at 5:33
So far no answers written out in English. Isn't the crux of the matter that if there were no $x$ such that $P(x)$, then $P(x)$ would be false for all $x$ and thus would imply absolutely anything?
– Wildcard
Nov 26 at 5:33
add a comment |
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
Resolved! I realised that I was going nowhere by assuming the opposite of what I was given as premise... I obviously had to assume the opposite of what I was trying to prove:
answered Nov 25 at 20:57
35308
827
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 at 22:38
add a comment |
up vote
2
down vote
1) ¬∀x(P(x) → Q(x)) --- premise
2) ¬∃xP(x) --- assumed [a]
3) P(x) --- assumed [b]
4) ∃xP(x) --- from 3) by ∃-intro
5) ⊥ --- contradiction : from 2) and 4)
6) Q(x) --- from 5) by ⊥-elim
7) P(x) → Q(x) --- from 3) and 6) by →-intro, discharging [b]
8) ∀x(P(x) → Q(x)) --- from 7) by ∀-intro
9) ⊥ --- contradiction : from 1) and 8)
10) ∃xP(x) --- from 2) by Double Negation (or ¬-elim), discharging [a].
answered Nov 25 at 20:07
Mauro ALLEGRANZA
27k21860
add a comment |
up vote
2
down vote
The following proof is the same as Mauro ALLEGRANZA's but it uses Klement's Fitch-style proof checker. Descriptions of the rules are in forallx. Both are available online and listed below. They may help as supplementary material to what you are currently using.
You may be required in your proof checker to represent contradictions as conjunctions of contradictory statements. This proof checker only requires noting the contradiction as "⊥" and listing the contradictory lines such as I did on lines 5 and 9.
References
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 Nov 25 at 20:36
Frank Hubeny
6,09641244
I managed to do it, but my result is slightly different from yours. I tried yours out of curiosity, and it does not work in the Fitch program I have. I'm puzzled by the fact that your proof checker allowed you to export the temporary assumption 'a' outside of its subproof!
– 35308
Nov 25 at 21:07
@35308 You could look at that subproof (lines 3-6) as saying the same thing as the conditional statement on line 7 "Pa > Qa". Given the assumption Pa, I can derive Qa with what is available (namely line 2). The subproof just showed how that was done and the proof checker required that I show that before using line 7.
– Frank Hubeny
Nov 25 at 21:21
The issue is that you really should not be allowed to introduce a new free term 'a' without raising a new context to isolate it.
– Graham Kemp
Nov 25 at 22:48
add a comment |
up vote
1
down vote
Pr. ~∀x(P(x)->Q(x))
2.∃x~(P(x)->Q(x)) ~ Universal out Pr.
3.~(P(a)->Q(a)) Existential out (x/a) 2
4.P(a)&~Q(a) ~ conditional out 3
5.P(a) Conjunction out 4
6.∃xP(x) Existential In 5
answered Nov 25 at 23:02
Bertrand Wittgenstein's Ghost
2617
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Resolved! I realised that I was going nowhere by assuming the opposite of what I was given as premise... I obviously had to assume the opposite of what I was trying to prove:
answered Nov 25 at 20:57
35308
827
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 at 22:38
add a comment |
up vote
3
down vote
accepted
Resolved! I realised that I was going nowhere by assuming the opposite of what I was given as premise... I obviously had to assume the opposite of what I was trying to prove:
answered Nov 25 at 20:57
35308
827
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 at 22:38
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Resolved! I realised that I was going nowhere by assuming the opposite of what I was given as premise... I obviously had to assume the opposite of what I was trying to prove:
answered Nov 25 at 20:57
35308
827
Resolved! I realised that I was going nowhere by assuming the opposite of what I was given as premise... I obviously had to assume the opposite of what I was trying to prove:
answered Nov 25 at 20:57
35308
827
answered Nov 25 at 20:57
35308
827
answered Nov 25 at 20:57
35308
827
answered Nov 25 at 20:57
35308
827
827
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 at 22:38
add a comment |
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 at 22:38
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 at 22:38
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 at 22:38
add a comment |
up vote
2
down vote
1) ¬∀x(P(x) → Q(x)) --- premise
2) ¬∃xP(x) --- assumed [a]
3) P(x) --- assumed [b]
4) ∃xP(x) --- from 3) by ∃-intro
5) ⊥ --- contradiction : from 2) and 4)
6) Q(x) --- from 5) by ⊥-elim
7) P(x) → Q(x) --- from 3) and 6) by →-intro, discharging [b]
8) ∀x(P(x) → Q(x)) --- from 7) by ∀-intro
9) ⊥ --- contradiction : from 1) and 8)
10) ∃xP(x) --- from 2) by Double Negation (or ¬-elim), discharging [a].
answered Nov 25 at 20:07
Mauro ALLEGRANZA
27k21860
add a comment |
up vote
2
down vote
1) ¬∀x(P(x) → Q(x)) --- premise
2) ¬∃xP(x) --- assumed [a]
3) P(x) --- assumed [b]
4) ∃xP(x) --- from 3) by ∃-intro
5) ⊥ --- contradiction : from 2) and 4)
6) Q(x) --- from 5) by ⊥-elim
7) P(x) → Q(x) --- from 3) and 6) by →-intro, discharging [b]
8) ∀x(P(x) → Q(x)) --- from 7) by ∀-intro
9) ⊥ --- contradiction : from 1) and 8)
10) ∃xP(x) --- from 2) by Double Negation (or ¬-elim), discharging [a].
answered Nov 25 at 20:07
Mauro ALLEGRANZA
27k21860
add a comment |
up vote
2
down vote
up vote
2
down vote
1) ¬∀x(P(x) → Q(x)) --- premise
2) ¬∃xP(x) --- assumed [a]
3) P(x) --- assumed [b]
4) ∃xP(x) --- from 3) by ∃-intro
5) ⊥ --- contradiction : from 2) and 4)
6) Q(x) --- from 5) by ⊥-elim
7) P(x) → Q(x) --- from 3) and 6) by →-intro, discharging [b]
8) ∀x(P(x) → Q(x)) --- from 7) by ∀-intro
9) ⊥ --- contradiction : from 1) and 8)
10) ∃xP(x) --- from 2) by Double Negation (or ¬-elim), discharging [a].
answered Nov 25 at 20:07
Mauro ALLEGRANZA
27k21860
1) ¬∀x(P(x) → Q(x)) --- premise
2) ¬∃xP(x) --- assumed [a]
3) P(x) --- assumed [b]
4) ∃xP(x) --- from 3) by ∃-intro
5) ⊥ --- contradiction : from 2) and 4)
6) Q(x) --- from 5) by ⊥-elim
7) P(x) → Q(x) --- from 3) and 6) by →-intro, discharging [b]
8) ∀x(P(x) → Q(x)) --- from 7) by ∀-intro
9) ⊥ --- contradiction : from 1) and 8)
10) ∃xP(x) --- from 2) by Double Negation (or ¬-elim), discharging [a].
answered Nov 25 at 20:07
Mauro ALLEGRANZA
27k21860
edited Nov 25 at 20:21
answered Nov 25 at 20:07
Mauro ALLEGRANZA
27k21860
answered Nov 25 at 20:07
Mauro ALLEGRANZA
27k21860
answered Nov 25 at 20:07
Mauro ALLEGRANZA
27k21860
27k21860
add a comment |
add a comment |
up vote
2
down vote
The following proof is the same as Mauro ALLEGRANZA's but it uses Klement's Fitch-style proof checker. Descriptions of the rules are in forallx. Both are available online and listed below. They may help as supplementary material to what you are currently using.
You may be required in your proof checker to represent contradictions as conjunctions of contradictory statements. This proof checker only requires noting the contradiction as "⊥" and listing the contradictory lines such as I did on lines 5 and 9.
References
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 Nov 25 at 20:36
Frank Hubeny
6,09641244
I managed to do it, but my result is slightly different from yours. I tried yours out of curiosity, and it does not work in the Fitch program I have. I'm puzzled by the fact that your proof checker allowed you to export the temporary assumption 'a' outside of its subproof!
– 35308
Nov 25 at 21:07
@35308 You could look at that subproof (lines 3-6) as saying the same thing as the conditional statement on line 7 "Pa > Qa". Given the assumption Pa, I can derive Qa with what is available (namely line 2). The subproof just showed how that was done and the proof checker required that I show that before using line 7.
– Frank Hubeny
Nov 25 at 21:21
The issue is that you really should not be allowed to introduce a new free term 'a' without raising a new context to isolate it.
– Graham Kemp
Nov 25 at 22:48
add a comment |
up vote
2
down vote
The following proof is the same as Mauro ALLEGRANZA's but it uses Klement's Fitch-style proof checker. Descriptions of the rules are in forallx. Both are available online and listed below. They may help as supplementary material to what you are currently using.
You may be required in your proof checker to represent contradictions as conjunctions of contradictory statements. This proof checker only requires noting the contradiction as "⊥" and listing the contradictory lines such as I did on lines 5 and 9.
References
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 Nov 25 at 20:36
Frank Hubeny
6,09641244
I managed to do it, but my result is slightly different from yours. I tried yours out of curiosity, and it does not work in the Fitch program I have. I'm puzzled by the fact that your proof checker allowed you to export the temporary assumption 'a' outside of its subproof!
– 35308
Nov 25 at 21:07
@35308 You could look at that subproof (lines 3-6) as saying the same thing as the conditional statement on line 7 "Pa > Qa". Given the assumption Pa, I can derive Qa with what is available (namely line 2). The subproof just showed how that was done and the proof checker required that I show that before using line 7.
– Frank Hubeny
Nov 25 at 21:21
The issue is that you really should not be allowed to introduce a new free term 'a' without raising a new context to isolate it.
– Graham Kemp
Nov 25 at 22:48
add a comment |
up vote
2
down vote
up vote
2
down vote
The following proof is the same as Mauro ALLEGRANZA's but it uses Klement's Fitch-style proof checker. Descriptions of the rules are in forallx. Both are available online and listed below. They may help as supplementary material to what you are currently using.
You may be required in your proof checker to represent contradictions as conjunctions of contradictory statements. This proof checker only requires noting the contradiction as "⊥" and listing the contradictory lines such as I did on lines 5 and 9.
References
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 Nov 25 at 20:36
Frank Hubeny
6,09641244
The following proof is the same as Mauro ALLEGRANZA's but it uses Klement's Fitch-style proof checker. Descriptions of the rules are in forallx. Both are available online and listed below. They may help as supplementary material to what you are currently using.
You may be required in your proof checker to represent contradictions as conjunctions of contradictory statements. This proof checker only requires noting the contradiction as "⊥" and listing the contradictory lines such as I did on lines 5 and 9.
References
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 Nov 25 at 20:36
Frank Hubeny
6,09641244
answered Nov 25 at 20:36
Frank Hubeny
6,09641244
answered Nov 25 at 20:36
Frank Hubeny
6,09641244
answered Nov 25 at 20:36
Frank Hubeny
6,09641244
6,09641244
I managed to do it, but my result is slightly different from yours. I tried yours out of curiosity, and it does not work in the Fitch program I have. I'm puzzled by the fact that your proof checker allowed you to export the temporary assumption 'a' outside of its subproof!
– 35308
Nov 25 at 21:07
@35308 You could look at that subproof (lines 3-6) as saying the same thing as the conditional statement on line 7 "Pa > Qa". Given the assumption Pa, I can derive Qa with what is available (namely line 2). The subproof just showed how that was done and the proof checker required that I show that before using line 7.
– Frank Hubeny
Nov 25 at 21:21
The issue is that you really should not be allowed to introduce a new free term 'a' without raising a new context to isolate it.
– Graham Kemp
Nov 25 at 22:48
add a comment |
I managed to do it, but my result is slightly different from yours. I tried yours out of curiosity, and it does not work in the Fitch program I have. I'm puzzled by the fact that your proof checker allowed you to export the temporary assumption 'a' outside of its subproof!
– 35308
Nov 25 at 21:07
@35308 You could look at that subproof (lines 3-6) as saying the same thing as the conditional statement on line 7 "Pa > Qa". Given the assumption Pa, I can derive Qa with what is available (namely line 2). The subproof just showed how that was done and the proof checker required that I show that before using line 7.
– Frank Hubeny
Nov 25 at 21:21
The issue is that you really should not be allowed to introduce a new free term 'a' without raising a new context to isolate it.
– Graham Kemp
Nov 25 at 22:48
I managed to do it, but my result is slightly different from yours. I tried yours out of curiosity, and it does not work in the Fitch program I have. I'm puzzled by the fact that your proof checker allowed you to export the temporary assumption 'a' outside of its subproof!
– 35308
Nov 25 at 21:07
I managed to do it, but my result is slightly different from yours. I tried yours out of curiosity, and it does not work in the Fitch program I have. I'm puzzled by the fact that your proof checker allowed you to export the temporary assumption 'a' outside of its subproof!
– 35308
Nov 25 at 21:07
@35308 You could look at that subproof (lines 3-6) as saying the same thing as the conditional statement on line 7 "Pa > Qa". Given the assumption Pa, I can derive Qa with what is available (namely line 2). The subproof just showed how that was done and the proof checker required that I show that before using line 7.
– Frank Hubeny
Nov 25 at 21:21
@35308 You could look at that subproof (lines 3-6) as saying the same thing as the conditional statement on line 7 "Pa > Qa". Given the assumption Pa, I can derive Qa with what is available (namely line 2). The subproof just showed how that was done and the proof checker required that I show that before using line 7.
– Frank Hubeny
Nov 25 at 21:21
The issue is that you really should not be allowed to introduce a new free term 'a' without raising a new context to isolate it.
– Graham Kemp
Nov 25 at 22:48
The issue is that you really should not be allowed to introduce a new free term 'a' without raising a new context to isolate it.
– Graham Kemp
Nov 25 at 22:48
add a comment |
up vote
1
down vote
Pr. ~∀x(P(x)->Q(x))
2.∃x~(P(x)->Q(x)) ~ Universal out Pr.
3.~(P(a)->Q(a)) Existential out (x/a) 2
4.P(a)&~Q(a) ~ conditional out 3
5.P(a) Conjunction out 4
6.∃xP(x) Existential In 5
answered Nov 25 at 23:02
Bertrand Wittgenstein's Ghost
2617
add a comment |
up vote
1
down vote
Pr. ~∀x(P(x)->Q(x))
2.∃x~(P(x)->Q(x)) ~ Universal out Pr.
3.~(P(a)->Q(a)) Existential out (x/a) 2
4.P(a)&~Q(a) ~ conditional out 3
5.P(a) Conjunction out 4
6.∃xP(x) Existential In 5
answered Nov 25 at 23:02
Bertrand Wittgenstein's Ghost
2617
add a comment |
up vote
1
down vote
up vote
1
down vote
Pr. ~∀x(P(x)->Q(x))
2.∃x~(P(x)->Q(x)) ~ Universal out Pr.
3.~(P(a)->Q(a)) Existential out (x/a) 2
4.P(a)&~Q(a) ~ conditional out 3
5.P(a) Conjunction out 4
6.∃xP(x) Existential In 5
answered Nov 25 at 23:02
Bertrand Wittgenstein's Ghost
2617
Pr. ~∀x(P(x)->Q(x))
2.∃x~(P(x)->Q(x)) ~ Universal out Pr.
3.~(P(a)->Q(a)) Existential out (x/a) 2
4.P(a)&~Q(a) ~ conditional out 3
5.P(a) Conjunction out 4
6.∃xP(x) Existential In 5
answered Nov 25 at 23:02
Bertrand Wittgenstein's Ghost
2617
answered Nov 25 at 23:02
Bertrand Wittgenstein's Ghost
2617
answered Nov 25 at 23:02
Bertrand Wittgenstein's Ghost
2617
answered Nov 25 at 23:02
Bertrand Wittgenstein's Ghost
2617
2617
add a comment |
add a comment |
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So far no answers written out in English. Isn't the crux of the matter that if there were no $x$ such that $P(x)$, then $P(x)$ would be false for all $x$ and thus would imply absolutely anything?
– Wildcard
Nov 26 at 5:33