How to prove ~($forall$x Q(x)) is logically equivalent to $exists$x(~Q(x)) using natural deduction for first...











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I am thinking of assuming Q(x1) and then deriving to reach to a contradiction but I have not been able to do so.






logic first-order-logic natural-deduction







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  • Could you give us some more background? For instance, in my book, this is the definition of $forall$.
    – Arthur
    Nov 21 at 12:43












  • here $forall$ stand for "for all" x that belong to domain of discourse.
    – user10143594
    Nov 21 at 12:51















up vote
0
down vote

favorite












I am thinking of assuming Q(x1) and then deriving to reach to a contradiction but I have not been able to do so.






logic first-order-logic natural-deduction







share|cite|improve this question
























  • Could you give us some more background? For instance, in my book, this is the definition of $forall$.
    – Arthur
    Nov 21 at 12:43












  • here $forall$ stand for "for all" x that belong to domain of discourse.
    – user10143594
    Nov 21 at 12:51













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am thinking of assuming Q(x1) and then deriving to reach to a contradiction but I have not been able to do so.






logic first-order-logic natural-deduction







share|cite|improve this question















I am thinking of assuming Q(x1) and then deriving to reach to a contradiction but I have not been able to do so.






logic first-order-logic natural-deduction




logic first-order-logic natural-deduction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 12:45









Mauro ALLEGRANZA

63.8k448110




63.8k448110










asked Nov 21 at 12:41









user10143594

103




103












  • Could you give us some more background? For instance, in my book, this is the definition of $forall$.
    – Arthur
    Nov 21 at 12:43












  • here $forall$ stand for "for all" x that belong to domain of discourse.
    – user10143594
    Nov 21 at 12:51


















  • Could you give us some more background? For instance, in my book, this is the definition of $forall$.
    – Arthur
    Nov 21 at 12:43












  • here $forall$ stand for "for all" x that belong to domain of discourse.
    – user10143594
    Nov 21 at 12:51
















Could you give us some more background? For instance, in my book, this is the definition of $forall$.
– Arthur
Nov 21 at 12:43






Could you give us some more background? For instance, in my book, this is the definition of $forall$.
– Arthur
Nov 21 at 12:43














here $forall$ stand for "for all" x that belong to domain of discourse.
– user10143594
Nov 21 at 12:51




here $forall$ stand for "for all" x that belong to domain of discourse.
– user10143594
Nov 21 at 12:51










1 Answer
1






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up vote
0
down vote



accepted










Hint



1st part, using Natural Deduction :



1) $exists x lnot Qx$ --- premise



2) $forall x Qx$ --- assumed [a]



3) $lnot Qa$ --- assumed [b] from 1) by $exists$-elim



4) $Qa$ --- from 2) by $forall$-elim



5) $bot$ --- contradiction : from 3) and 4)



6) $bot$ --- from 1) and 3) and 5) by $exists$-elim, discharging [b]




7) $lnot forall x Qx$ --- from 2) and 6) by $lnot$-intro, discharging [a].





2nd part : quite similar, using Double Negation.



1) $lnot forall x Qx$ --- premise



2) $lnot exists x lnot Qx$ --- assumed [a]



3) $lnot Qx$ --- assumed [b]



and so on ...






share|cite|improve this answer























  • Thanks. Can you also show for other way round?
    – user10143594
    Nov 21 at 12:46











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Hint



1st part, using Natural Deduction :



1) $exists x lnot Qx$ --- premise



2) $forall x Qx$ --- assumed [a]



3) $lnot Qa$ --- assumed [b] from 1) by $exists$-elim



4) $Qa$ --- from 2) by $forall$-elim



5) $bot$ --- contradiction : from 3) and 4)



6) $bot$ --- from 1) and 3) and 5) by $exists$-elim, discharging [b]




7) $lnot forall x Qx$ --- from 2) and 6) by $lnot$-intro, discharging [a].





2nd part : quite similar, using Double Negation.



1) $lnot forall x Qx$ --- premise



2) $lnot exists x lnot Qx$ --- assumed [a]



3) $lnot Qx$ --- assumed [b]



and so on ...






share|cite|improve this answer























  • Thanks. Can you also show for other way round?
    – user10143594
    Nov 21 at 12:46















up vote
0
down vote



accepted










Hint



1st part, using Natural Deduction :



1) $exists x lnot Qx$ --- premise



2) $forall x Qx$ --- assumed [a]



3) $lnot Qa$ --- assumed [b] from 1) by $exists$-elim



4) $Qa$ --- from 2) by $forall$-elim



5) $bot$ --- contradiction : from 3) and 4)



6) $bot$ --- from 1) and 3) and 5) by $exists$-elim, discharging [b]




7) $lnot forall x Qx$ --- from 2) and 6) by $lnot$-intro, discharging [a].





2nd part : quite similar, using Double Negation.



1) $lnot forall x Qx$ --- premise



2) $lnot exists x lnot Qx$ --- assumed [a]



3) $lnot Qx$ --- assumed [b]



and so on ...






share|cite|improve this answer























  • Thanks. Can you also show for other way round?
    – user10143594
    Nov 21 at 12:46













up vote
0
down vote



accepted







up vote
0
down vote



accepted






Hint



1st part, using Natural Deduction :



1) $exists x lnot Qx$ --- premise



2) $forall x Qx$ --- assumed [a]



3) $lnot Qa$ --- assumed [b] from 1) by $exists$-elim



4) $Qa$ --- from 2) by $forall$-elim



5) $bot$ --- contradiction : from 3) and 4)



6) $bot$ --- from 1) and 3) and 5) by $exists$-elim, discharging [b]




7) $lnot forall x Qx$ --- from 2) and 6) by $lnot$-intro, discharging [a].





2nd part : quite similar, using Double Negation.



1) $lnot forall x Qx$ --- premise



2) $lnot exists x lnot Qx$ --- assumed [a]



3) $lnot Qx$ --- assumed [b]



and so on ...






share|cite|improve this answer














Hint



1st part, using Natural Deduction :



1) $exists x lnot Qx$ --- premise



2) $forall x Qx$ --- assumed [a]



3) $lnot Qa$ --- assumed [b] from 1) by $exists$-elim



4) $Qa$ --- from 2) by $forall$-elim



5) $bot$ --- contradiction : from 3) and 4)



6) $bot$ --- from 1) and 3) and 5) by $exists$-elim, discharging [b]




7) $lnot forall x Qx$ --- from 2) and 6) by $lnot$-intro, discharging [a].





2nd part : quite similar, using Double Negation.



1) $lnot forall x Qx$ --- premise



2) $lnot exists x lnot Qx$ --- assumed [a]



3) $lnot Qx$ --- assumed [b]



and so on ...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 13:14

























answered Nov 21 at 12:45









Mauro ALLEGRANZA

63.8k448110




63.8k448110












  • Thanks. Can you also show for other way round?
    – user10143594
    Nov 21 at 12:46


















  • Thanks. Can you also show for other way round?
    – user10143594
    Nov 21 at 12:46
















Thanks. Can you also show for other way round?
– user10143594
Nov 21 at 12:46




Thanks. Can you also show for other way round?
– user10143594
Nov 21 at 12:46


















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