How to prove ~($forall$x Q(x)) is logically equivalent to $exists$x(~Q(x)) using natural deduction for first...
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
edited Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 12:41
user10143594
103
add a comment |
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
edited Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 12:41
user10143594
103
Could you give us some more background? For instance, in my book, this is the definition of $forall$.
– Arthur
Nov 21 '18 at 12:43
here $forall$ stand for "for all" x that belong to domain of discourse.
– user10143594
Nov 21 '18 at 12:51
add a comment |
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
edited Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 12:41
user10143594
103
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
edited Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 12:41
user10143594
103
edited Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 12:41
user10143594
103
edited Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
edited Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
edited Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
64.4k448112
asked Nov 21 '18 at 12:41
user10143594
103
asked Nov 21 '18 at 12:41
user10143594
103
asked Nov 21 '18 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 '18 at 12:43
here $forall$ stand for "for all" x that belong to domain of discourse.
– user10143594
Nov 21 '18 at 12:51
add a comment |
Could you give us some more background? For instance, in my book, this is the definition of $forall$.
– Arthur
Nov 21 '18 at 12:43
here $forall$ stand for "for all" x that belong to domain of discourse.
– user10143594
Nov 21 '18 at 12:51
Could you give us some more background? For instance, in my book, this is the definition of $forall$.
– Arthur
Nov 21 '18 at 12:43
Could you give us some more background? For instance, in my book, this is the definition of $forall$.
– Arthur
Nov 21 '18 at 12:43
here $forall$ stand for "for all" x that belong to domain of discourse.
– user10143594
Nov 21 '18 at 12:51
here $forall$ stand for "for all" x that belong to domain of discourse.
– user10143594
Nov 21 '18 at 12:51
add a comment |
1 Answer
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oldest
votes
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 ...
answered Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
Thanks. Can you also show for other way round?
– user10143594
Nov 21 '18 at 12:46
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 ...
answered Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
Thanks. Can you also show for other way round?
– user10143594
Nov 21 '18 at 12:46
add a comment |
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 ...
answered Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
Thanks. Can you also show for other way round?
– user10143594
Nov 21 '18 at 12:46
add a comment |
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 ...
answered Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
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 ...
answered Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
edited Nov 21 '18 at 13:14
answered Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
answered Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
answered Nov 21 '18 at 12:45
Mauro ALLEGRANZA
64.4k448112
64.4k448112
Thanks. Can you also show for other way round?
– user10143594
Nov 21 '18 at 12:46
add a comment |
Thanks. Can you also show for other way round?
– user10143594
Nov 21 '18 at 12:46
Thanks. Can you also show for other way round?
– user10143594
Nov 21 '18 at 12:46
Thanks. Can you also show for other way round?
– user10143594
Nov 21 '18 at 12:46
add a comment |
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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 '18 at 12:43
here $forall$ stand for "for all" x that belong to domain of discourse.
– user10143594
Nov 21 '18 at 12:51