How to prove ‘∃xP(x)’ from ‘¬∀x(P(x)→Q(x))’
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 '18 at 19:56
3530835308
827
add a comment |
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 '18 at 19:56
3530835308
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 '18 at 5:33
add a comment |
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 '18 at 19:56
3530835308
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 '18 at 19:56
3530835308
827
asked Nov 25 '18 at 19:56
3530835308
827
edited Nov 25 '18 at 20:58
35308
asked Nov 25 '18 at 19:56
3530835308
827
asked Nov 25 '18 at 19:56
3530835308
827
asked Nov 25 '18 at 19:56
3530835308
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 '18 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 '18 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 '18 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 '18 at 5:33
add a comment |
4 Answers
4
active
oldest
votes
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 '18 at 20:57
3530835308
827
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 '18 at 22:38
add a comment |
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 '18 at 20:07
Mauro ALLEGRANZAMauro ALLEGRANZA
28.3k21963
add a comment |
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 '18 at 20:36
Frank HubenyFrank Hubeny
7,87251447
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 '18 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 '18 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 '18 at 22:48
add a comment |
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 '18 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
3238
add a comment |
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4 Answers
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4 Answers
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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 '18 at 20:57
3530835308
827
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 '18 at 22:38
add a comment |
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 '18 at 20:57
3530835308
827
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 '18 at 22:38
add a comment |
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 '18 at 20:57
3530835308
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 '18 at 20:57
3530835308
827
answered Nov 25 '18 at 20:57
3530835308
827
answered Nov 25 '18 at 20:57
3530835308
827
answered Nov 25 '18 at 20:57
3530835308
827
827
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 '18 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 '18 at 22:38
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 '18 at 22:38
Indeed. You have a correct solution. You can now accept your own answer to close this question.
– Graham Kemp
Nov 25 '18 at 22:38
add a comment |
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 '18 at 20:07
Mauro ALLEGRANZAMauro ALLEGRANZA
28.3k21963
add a comment |
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 '18 at 20:07
Mauro ALLEGRANZAMauro ALLEGRANZA
28.3k21963
add a comment |
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 '18 at 20:07
Mauro ALLEGRANZAMauro ALLEGRANZA
28.3k21963
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 '18 at 20:07
Mauro ALLEGRANZAMauro ALLEGRANZA
28.3k21963
edited Nov 25 '18 at 20:21
answered Nov 25 '18 at 20:07
Mauro ALLEGRANZAMauro ALLEGRANZA
28.3k21963
answered Nov 25 '18 at 20:07
Mauro ALLEGRANZAMauro ALLEGRANZA
28.3k21963
answered Nov 25 '18 at 20:07
Mauro ALLEGRANZAMauro ALLEGRANZA
28.3k21963
28.3k21963
add a comment |
add a comment |
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 '18 at 20:36
Frank HubenyFrank Hubeny
7,87251447
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 '18 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 '18 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 '18 at 22:48
add a comment |
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 '18 at 20:36
Frank HubenyFrank Hubeny
7,87251447
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 '18 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 '18 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 '18 at 22:48
add a comment |
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 '18 at 20:36
Frank HubenyFrank Hubeny
7,87251447
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 '18 at 20:36
Frank HubenyFrank Hubeny
7,87251447
answered Nov 25 '18 at 20:36
Frank HubenyFrank Hubeny
7,87251447
answered Nov 25 '18 at 20:36
Frank HubenyFrank Hubeny
7,87251447
answered Nov 25 '18 at 20:36
Frank HubenyFrank Hubeny
7,87251447
7,87251447
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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 22:48
add a comment |
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 '18 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
3238
add a comment |
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 '18 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
3238
add a comment |
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 '18 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
3238
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 '18 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
3238
answered Nov 25 '18 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
3238
answered Nov 25 '18 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
3238
answered Nov 25 '18 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
3238
3238
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 '18 at 5:33