Mixing
Tricky proof in Natural Deduction [¬∀x∃y¬Rxy ⊢ ∃x∀yRxy]-help appreciated
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As it says in the title, I'm trying to prove the statement ¬∀x∃y¬Rxy ⊢ ∃x∀yRxy in the system of Natural Deduction, with basic rules for negation intro and elimination, and existential or universal quantifier intro and elim.
I can use the premiss to create a contradiction, so the penultimate line might be:
¬∀x∃y¬Rxy ∀x∃y¬Rxy
∃x∀yRxy
But how do I derive the assumption ∀x∃y¬Rxy? It might be:
¬Rab
∃y¬Ray
∀x∃y¬Rxy
So ¬Rab must be derived from some assumption Rab that leads to contradiction-that's where I'm stuck! Any help appreciated-this is not my homework or anything, it just bugs me!
predicate-logic first-order-logic natural-deduction
asked Nov 18 '15 at 20:35
user211309user211309
414
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add a comment |
$begingroup$
As it says in the title, I'm trying to prove the statement ¬∀x∃y¬Rxy ⊢ ∃x∀yRxy in the system of Natural Deduction, with basic rules for negation intro and elimination, and existential or universal quantifier intro and elim.
I can use the premiss to create a contradiction, so the penultimate line might be:
¬∀x∃y¬Rxy ∀x∃y¬Rxy
∃x∀yRxy
But how do I derive the assumption ∀x∃y¬Rxy? It might be:
¬Rab
∃y¬Ray
∀x∃y¬Rxy
So ¬Rab must be derived from some assumption Rab that leads to contradiction-that's where I'm stuck! Any help appreciated-this is not my homework or anything, it just bugs me!
predicate-logic first-order-logic natural-deduction
asked Nov 18 '15 at 20:35
user211309user211309
414
$endgroup$
add a comment |
$begingroup$
As it says in the title, I'm trying to prove the statement ¬∀x∃y¬Rxy ⊢ ∃x∀yRxy in the system of Natural Deduction, with basic rules for negation intro and elimination, and existential or universal quantifier intro and elim.
I can use the premiss to create a contradiction, so the penultimate line might be:
¬∀x∃y¬Rxy ∀x∃y¬Rxy
∃x∀yRxy
But how do I derive the assumption ∀x∃y¬Rxy? It might be:
¬Rab
∃y¬Ray
∀x∃y¬Rxy
So ¬Rab must be derived from some assumption Rab that leads to contradiction-that's where I'm stuck! Any help appreciated-this is not my homework or anything, it just bugs me!
predicate-logic first-order-logic natural-deduction
asked Nov 18 '15 at 20:35
user211309user211309
414
$endgroup$
As it says in the title, I'm trying to prove the statement ¬∀x∃y¬Rxy ⊢ ∃x∀yRxy in the system of Natural Deduction, with basic rules for negation intro and elimination, and existential or universal quantifier intro and elim.
I can use the premiss to create a contradiction, so the penultimate line might be:
¬∀x∃y¬Rxy ∀x∃y¬Rxy
∃x∀yRxy
But how do I derive the assumption ∀x∃y¬Rxy? It might be:
¬Rab
∃y¬Ray
∀x∃y¬Rxy
So ¬Rab must be derived from some assumption Rab that leads to contradiction-that's where I'm stuck! Any help appreciated-this is not my homework or anything, it just bugs me!
predicate-logic first-order-logic natural-deduction
predicate-logic first-order-logic natural-deduction
asked Nov 18 '15 at 20:35
user211309user211309
414
asked Nov 18 '15 at 20:35
user211309user211309
414
asked Nov 18 '15 at 20:35
user211309user211309
414
asked Nov 18 '15 at 20:35
user211309user211309
414
asked Nov 18 '15 at 20:35
user211309user211309
414
414
add a comment |
add a comment |
1 Answer
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$begingroup$
We need :
1) $¬∀x∃y¬Rxy$ --- premise
2) $¬∃x∀yRxy$ --- assumption [a] : the assumption to be contradicted, deriving (by Double Negation) the sought conclusion : $∃x∀yRxy$
3) $∀yRxy$ --- assumption [b] : in order to derive a contradiction with 2) to conclude, by $¬$-intro, with : $¬∀yRxy$, discharging [b].
Now we need a "detour", in order to derive from $¬∀yRxy$ the classically equivalent : $∃y¬Rxy$.
Having derived $∃y¬Rxy$, we can "generalize" it to $∀x∃y¬Rxy$. We can do it, because neither in the premise nor into assumption [a] $x$ is free.
Thus, $∀x∃y¬Rxy$ contradicts the premise 1) and we can conclude by Double Negation again with :
$∃x∀yRxy$,
discharging [a].
answered Nov 19 '15 at 8:00
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
$endgroup$
$begingroup$
Thanks Mauro, I've got it now! I didn't use the right assumption, that got me into trouble.
$endgroup$
– user211309
Nov 19 '15 at 10:43
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We need :
1) $¬∀x∃y¬Rxy$ --- premise
2) $¬∃x∀yRxy$ --- assumption [a] : the assumption to be contradicted, deriving (by Double Negation) the sought conclusion : $∃x∀yRxy$
3) $∀yRxy$ --- assumption [b] : in order to derive a contradiction with 2) to conclude, by $¬$-intro, with : $¬∀yRxy$, discharging [b].
Now we need a "detour", in order to derive from $¬∀yRxy$ the classically equivalent : $∃y¬Rxy$.
Having derived $∃y¬Rxy$, we can "generalize" it to $∀x∃y¬Rxy$. We can do it, because neither in the premise nor into assumption [a] $x$ is free.
Thus, $∀x∃y¬Rxy$ contradicts the premise 1) and we can conclude by Double Negation again with :
$∃x∀yRxy$,
discharging [a].
answered Nov 19 '15 at 8:00
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
$endgroup$
$begingroup$
Thanks Mauro, I've got it now! I didn't use the right assumption, that got me into trouble.
$endgroup$
– user211309
Nov 19 '15 at 10:43
add a comment |
$begingroup$
We need :
1) $¬∀x∃y¬Rxy$ --- premise
2) $¬∃x∀yRxy$ --- assumption [a] : the assumption to be contradicted, deriving (by Double Negation) the sought conclusion : $∃x∀yRxy$
3) $∀yRxy$ --- assumption [b] : in order to derive a contradiction with 2) to conclude, by $¬$-intro, with : $¬∀yRxy$, discharging [b].
Now we need a "detour", in order to derive from $¬∀yRxy$ the classically equivalent : $∃y¬Rxy$.
Having derived $∃y¬Rxy$, we can "generalize" it to $∀x∃y¬Rxy$. We can do it, because neither in the premise nor into assumption [a] $x$ is free.
Thus, $∀x∃y¬Rxy$ contradicts the premise 1) and we can conclude by Double Negation again with :
$∃x∀yRxy$,
discharging [a].
answered Nov 19 '15 at 8:00
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
$endgroup$
$begingroup$
Thanks Mauro, I've got it now! I didn't use the right assumption, that got me into trouble.
$endgroup$
– user211309
Nov 19 '15 at 10:43
add a comment |
$begingroup$
We need :
1) $¬∀x∃y¬Rxy$ --- premise
2) $¬∃x∀yRxy$ --- assumption [a] : the assumption to be contradicted, deriving (by Double Negation) the sought conclusion : $∃x∀yRxy$
3) $∀yRxy$ --- assumption [b] : in order to derive a contradiction with 2) to conclude, by $¬$-intro, with : $¬∀yRxy$, discharging [b].
Now we need a "detour", in order to derive from $¬∀yRxy$ the classically equivalent : $∃y¬Rxy$.
Having derived $∃y¬Rxy$, we can "generalize" it to $∀x∃y¬Rxy$. We can do it, because neither in the premise nor into assumption [a] $x$ is free.
Thus, $∀x∃y¬Rxy$ contradicts the premise 1) and we can conclude by Double Negation again with :
$∃x∀yRxy$,
discharging [a].
answered Nov 19 '15 at 8:00
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
$endgroup$
We need :
1) $¬∀x∃y¬Rxy$ --- premise
2) $¬∃x∀yRxy$ --- assumption [a] : the assumption to be contradicted, deriving (by Double Negation) the sought conclusion : $∃x∀yRxy$
3) $∀yRxy$ --- assumption [b] : in order to derive a contradiction with 2) to conclude, by $¬$-intro, with : $¬∀yRxy$, discharging [b].
Now we need a "detour", in order to derive from $¬∀yRxy$ the classically equivalent : $∃y¬Rxy$.
Having derived $∃y¬Rxy$, we can "generalize" it to $∀x∃y¬Rxy$. We can do it, because neither in the premise nor into assumption [a] $x$ is free.
Thus, $∀x∃y¬Rxy$ contradicts the premise 1) and we can conclude by Double Negation again with :
$∃x∀yRxy$,
discharging [a].
answered Nov 19 '15 at 8:00
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
answered Nov 19 '15 at 8:00
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
answered Nov 19 '15 at 8:00
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
answered Nov 19 '15 at 8:00
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
67.2k449115
$begingroup$
Thanks Mauro, I've got it now! I didn't use the right assumption, that got me into trouble.
$endgroup$
– user211309
Nov 19 '15 at 10:43
add a comment |
$begingroup$
Thanks Mauro, I've got it now! I didn't use the right assumption, that got me into trouble.
$endgroup$
– user211309
Nov 19 '15 at 10:43
$begingroup$
Thanks Mauro, I've got it now! I didn't use the right assumption, that got me into trouble.
$endgroup$
– user211309
Nov 19 '15 at 10:43
$begingroup$
Thanks Mauro, I've got it now! I didn't use the right assumption, that got me into trouble.
$endgroup$
– user211309
Nov 19 '15 at 10:43
add a comment |
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