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Given ¬(p ∨ q), use the Fitch system to prove (¬p ∧ ¬q).
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So, I understood that in order to succeed I should prove ¬p and ¬q and then use And Introduction.
However, I don't seem to figure out how to do this and was only able to prove (¬p ∨ ¬q) (I'm not sure if I'm even on the right path)
I hope somebody can help me out
logic propositional-calculus
asked Jul 27 at 15:20
Daniil Voloshin
41
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favorite
So, I understood that in order to succeed I should prove ¬p and ¬q and then use And Introduction.
However, I don't seem to figure out how to do this and was only able to prove (¬p ∨ ¬q) (I'm not sure if I'm even on the right path)
I hope somebody can help me out
logic propositional-calculus
asked Jul 27 at 15:20
Daniil Voloshin
41
1
Try assuming p, then get a contradiction, from which you can use the contradiction elimination rule.
– Adrian Keister
Jul 27 at 15:22
1
Well, there's more than one proof. Note that a proof is a particular sequence, so any change in order of steps makes a distinct proof. For one proof assume p first, get to $lnot$p. For another, assume q first and get to $lnot$q.
– Doug Spoonwood
Jul 27 at 16:44
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So, I understood that in order to succeed I should prove ¬p and ¬q and then use And Introduction.
However, I don't seem to figure out how to do this and was only able to prove (¬p ∨ ¬q) (I'm not sure if I'm even on the right path)
I hope somebody can help me out
logic propositional-calculus
asked Jul 27 at 15:20
Daniil Voloshin
41
So, I understood that in order to succeed I should prove ¬p and ¬q and then use And Introduction.
However, I don't seem to figure out how to do this and was only able to prove (¬p ∨ ¬q) (I'm not sure if I'm even on the right path)
I hope somebody can help me out
logic propositional-calculus
asked Jul 27 at 15:20
Daniil Voloshin
41
asked Jul 27 at 15:20
Daniil Voloshin
41
asked Jul 27 at 15:20
Daniil Voloshin
41
asked Jul 27 at 15:20
Daniil Voloshin
41
41
1
Try assuming p, then get a contradiction, from which you can use the contradiction elimination rule.
– Adrian Keister
Jul 27 at 15:22
1
Well, there's more than one proof. Note that a proof is a particular sequence, so any change in order of steps makes a distinct proof. For one proof assume p first, get to $lnot$p. For another, assume q first and get to $lnot$q.
– Doug Spoonwood
Jul 27 at 16:44
add a comment |Â
1
Try assuming p, then get a contradiction, from which you can use the contradiction elimination rule.
– Adrian Keister
Jul 27 at 15:22
1
Well, there's more than one proof. Note that a proof is a particular sequence, so any change in order of steps makes a distinct proof. For one proof assume p first, get to $lnot$p. For another, assume q first and get to $lnot$q.
– Doug Spoonwood
Jul 27 at 16:44
1
1
Try assuming p, then get a contradiction, from which you can use the contradiction elimination rule.
– Adrian Keister
Jul 27 at 15:22
Try assuming p, then get a contradiction, from which you can use the contradiction elimination rule.
– Adrian Keister
Jul 27 at 15:22
1
1
Well, there's more than one proof. Note that a proof is a particular sequence, so any change in order of steps makes a distinct proof. For one proof assume p first, get to $lnot$p. For another, assume q first and get to $lnot$q.
– Doug Spoonwood
Jul 27 at 16:44
Well, there's more than one proof. Note that a proof is a particular sequence, so any change in order of steps makes a distinct proof. For one proof assume p first, get to $lnot$p. For another, assume q first and get to $lnot$q.
– Doug Spoonwood
Jul 27 at 16:44
add a comment |Â
1 Answer
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This is one of De Morgan's laws
$lnotleft(plor qright)$
$pquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot p$
$qquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot q$
$lnot plandlnot q$
You just need to fill in the justification for each line.
answered Jul 27 at 16:49
Jon
500412
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
This is one of De Morgan's laws
$lnotleft(plor qright)$
$pquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot p$
$qquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot q$
$lnot plandlnot q$
You just need to fill in the justification for each line.
answered Jul 27 at 16:49
Jon
500412
add a comment |Â
up vote
2
down vote
This is one of De Morgan's laws
$lnotleft(plor qright)$
$pquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot p$
$qquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot q$
$lnot plandlnot q$
You just need to fill in the justification for each line.
answered Jul 27 at 16:49
Jon
500412
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is one of De Morgan's laws
$lnotleft(plor qright)$
$pquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot p$
$qquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot q$
$lnot plandlnot q$
You just need to fill in the justification for each line.
answered Jul 27 at 16:49
Jon
500412
This is one of De Morgan's laws
$lnotleft(plor qright)$
$pquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot p$
$qquad$ hyp.
$plor q$
$left(plor qright)landleft(lnotleft(plor qright)right)$
$lnot q$
$lnot plandlnot q$
You just need to fill in the justification for each line.
answered Jul 27 at 16:49
Jon
500412
edited Jul 27 at 16:58
answered Jul 27 at 16:49
Jon
500412
answered Jul 27 at 16:49
Jon
500412
answered Jul 27 at 16:49
Jon
500412
500412
add a comment |Â
add a comment |Â
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1
Try assuming p, then get a contradiction, from which you can use the contradiction elimination rule.
– Adrian Keister
Jul 27 at 15:22
1
Well, there's more than one proof. Note that a proof is a particular sequence, so any change in order of steps makes a distinct proof. For one proof assume p first, get to $lnot$p. For another, assume q first and get to $lnot$q.
– Doug Spoonwood
Jul 27 at 16:44