Mixing
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How to prove this logical equivalence in predicate logic?
$begingroup$
Prove that:
$ ( forall x)(forall y)(exists z)((P(x)Rightarrow Q(y)) wedge neg Q(z))$
is equivalent with
$ neg((exists xP(x) lor forall zQ(z))$
How should I attempt such problems?
I tried a lot like changing the implication in a disjunction. I applied Morgan rules here and there...
It looks like I should use the transitivity rule to get rid of the Y but I didn't find how
EDIT: Corrected question, thanks to Mauro
logic predicate-logic
edited Jan 30 at 16:34
Mauro ALLEGRANZA
67.7k449117
asked Jan 30 at 13:42
Ayoub RossiAyoub Rossi
11110
$endgroup$
add a comment |
$begingroup$
Prove that:
$ ( forall x)(forall y)(exists z)((P(x)Rightarrow Q(y)) wedge neg Q(z))$
is equivalent with
$ neg((exists xP(x) lor forall zQ(z))$
How should I attempt such problems?
I tried a lot like changing the implication in a disjunction. I applied Morgan rules here and there...
It looks like I should use the transitivity rule to get rid of the Y but I didn't find how
EDIT: Corrected question, thanks to Mauro
logic predicate-logic
edited Jan 30 at 16:34
Mauro ALLEGRANZA
67.7k449117
asked Jan 30 at 13:42
Ayoub RossiAyoub Rossi
11110
$endgroup$
$begingroup$
Is there an $y$ missing in the first formula ?
$endgroup$
– Mauro ALLEGRANZA
Jan 30 at 13:51
$begingroup$
Indeed, thanks!
$endgroup$
– Ayoub Rossi
Jan 30 at 13:54
$begingroup$
Well in the case that $Q(y)$ for all $y$, then it is true for any specific choice of $y$, namely $z$. Hence we can replace $forall y(P(x) rightarrow Q(y))$ with $ (P(x) rightarrow Q(z)) $
$endgroup$
– NazimJ
Jan 30 at 14:37
add a comment |
$begingroup$
Prove that:
$ ( forall x)(forall y)(exists z)((P(x)Rightarrow Q(y)) wedge neg Q(z))$
is equivalent with
$ neg((exists xP(x) lor forall zQ(z))$
How should I attempt such problems?
I tried a lot like changing the implication in a disjunction. I applied Morgan rules here and there...
It looks like I should use the transitivity rule to get rid of the Y but I didn't find how
EDIT: Corrected question, thanks to Mauro
logic predicate-logic
edited Jan 30 at 16:34
Mauro ALLEGRANZA
67.7k449117
asked Jan 30 at 13:42
Ayoub RossiAyoub Rossi
11110
$endgroup$
Prove that:
$ ( forall x)(forall y)(exists z)((P(x)Rightarrow Q(y)) wedge neg Q(z))$
is equivalent with
$ neg((exists xP(x) lor forall zQ(z))$
How should I attempt such problems?
I tried a lot like changing the implication in a disjunction. I applied Morgan rules here and there...
It looks like I should use the transitivity rule to get rid of the Y but I didn't find how
EDIT: Corrected question, thanks to Mauro
logic predicate-logic
logic predicate-logic
edited Jan 30 at 16:34
Mauro ALLEGRANZA
67.7k449117
asked Jan 30 at 13:42
Ayoub RossiAyoub Rossi
11110
edited Jan 30 at 16:34
Mauro ALLEGRANZA
67.7k449117
asked Jan 30 at 13:42
Ayoub RossiAyoub Rossi
11110
edited Jan 30 at 16:34
Mauro ALLEGRANZA
67.7k449117
edited Jan 30 at 16:34
Mauro ALLEGRANZA
67.7k449117
edited Jan 30 at 16:34
Mauro ALLEGRANZA
67.7k449117
67.7k449117
asked Jan 30 at 13:42
Ayoub RossiAyoub Rossi
11110
asked Jan 30 at 13:42
Ayoub RossiAyoub Rossi
11110
asked Jan 30 at 13:42
Ayoub RossiAyoub Rossi
11110
11110
$begingroup$
Is there an $y$ missing in the first formula ?
$endgroup$
– Mauro ALLEGRANZA
Jan 30 at 13:51
$begingroup$
Indeed, thanks!
$endgroup$
– Ayoub Rossi
Jan 30 at 13:54
$begingroup$
Well in the case that $Q(y)$ for all $y$, then it is true for any specific choice of $y$, namely $z$. Hence we can replace $forall y(P(x) rightarrow Q(y))$ with $ (P(x) rightarrow Q(z)) $
$endgroup$
– NazimJ
Jan 30 at 14:37
add a comment |
$begingroup$
Is there an $y$ missing in the first formula ?
$endgroup$
– Mauro ALLEGRANZA
Jan 30 at 13:51
$begingroup$
Indeed, thanks!
$endgroup$
– Ayoub Rossi
Jan 30 at 13:54
$begingroup$
Well in the case that $Q(y)$ for all $y$, then it is true for any specific choice of $y$, namely $z$. Hence we can replace $forall y(P(x) rightarrow Q(y))$ with $ (P(x) rightarrow Q(z)) $
$endgroup$
– NazimJ
Jan 30 at 14:37
$begingroup$
Is there an $y$ missing in the first formula ?
$endgroup$
– Mauro ALLEGRANZA
Jan 30 at 13:51
$begingroup$
Is there an $y$ missing in the first formula ?
$endgroup$
– Mauro ALLEGRANZA
Jan 30 at 13:51
$begingroup$
Indeed, thanks!
$endgroup$
– Ayoub Rossi
Jan 30 at 13:54
$begingroup$
Indeed, thanks!
$endgroup$
– Ayoub Rossi
Jan 30 at 13:54
$begingroup$
Well in the case that $Q(y)$ for all $y$, then it is true for any specific choice of $y$, namely $z$. Hence we can replace $forall y(P(x) rightarrow Q(y))$ with $ (P(x) rightarrow Q(z)) $
$endgroup$
– NazimJ
Jan 30 at 14:37
$begingroup$
Well in the case that $Q(y)$ for all $y$, then it is true for any specific choice of $y$, namely $z$. Hence we can replace $forall y(P(x) rightarrow Q(y))$ with $ (P(x) rightarrow Q(z)) $
$endgroup$
– NazimJ
Jan 30 at 14:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
We have to work with Prenex normal form equivalences.
The first formula is equivalent to :
$((∃x)P(x) to (∀y)Q(y)) ∧ (∃z)¬Q(z)$.
By De Morgan, the second formula is :
$¬(∃x) P(x) ∧ ¬(∀z)Q(z)$, i.e. $¬(∃x) P(x) ∧ (∃z)¬Q(z)$.
The first one implies the second : if we have that $(∃z)¬Q(z)$ holds, then it is false that $(∀y)Q(y)$ and thus also $(∃x)P(x)$ is false.
Thus : $¬(∃x) P(x)$ holds.
The second one implies the first : if $¬(∃x) P(x)$ holds, then $(∃x)P(x) to R$ holds, for $R$ whatever.
Thus : $(∃x)P(x) to (∀y)Q(y)$ holds.
answered Jan 30 at 16:34
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
$endgroup$
$begingroup$
One question: You say we have to work with prenex normal form but your first step switches to a non-prenex normal form if I'm not wrong?
$endgroup$
– Ayoub Rossi
Jan 30 at 17:04
$begingroup$
Second question: If I recognize a prenex normal form, should I always switch to an equivalent in non-prenex form
$endgroup$
– Ayoub Rossi
Jan 30 at 17:06
add a comment |
$begingroup$
$ forall x forall y exists z((P(x)rightarrow Q(y)) land neg Q(z)) overset{Prenex}{Leftrightarrow}$
$ forall x forall y ((P(x)rightarrow Q(y)) land exists z neg Q(z)) overset{Implication}{Leftrightarrow}$
$ forall x forall y ((neg P(x) lor Q(y)) land exists z neg Q(z))) overset{Distribution}{Leftrightarrow}$
$ forall x forall y ((neg P(x) land exists z neg Q(z)) lor ((Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ forall x (neg P(x) land exists z neg Q(z)) lor forall y (Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ ( forall x neg P(x) land exists z neg Q(z)) lor (forall y Q(y) land exists z neg Q(z)) overset{Quantifier Negation}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall z Q(z)) overset{Replacing Variables}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall y Q(y)) overset{Complement}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor bot overset{Identity}{Leftrightarrow}$
$ neg exists x P(x) land neg forall z Q(z)overset{DeMorgan}{Leftrightarrow}$
$ neg( exists x P(x) lor neg forall z Q(z))$
answered Jan 30 at 22:39
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
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$begingroup$
Hint
We have to work with Prenex normal form equivalences.
The first formula is equivalent to :
$((∃x)P(x) to (∀y)Q(y)) ∧ (∃z)¬Q(z)$.
By De Morgan, the second formula is :
$¬(∃x) P(x) ∧ ¬(∀z)Q(z)$, i.e. $¬(∃x) P(x) ∧ (∃z)¬Q(z)$.
The first one implies the second : if we have that $(∃z)¬Q(z)$ holds, then it is false that $(∀y)Q(y)$ and thus also $(∃x)P(x)$ is false.
Thus : $¬(∃x) P(x)$ holds.
The second one implies the first : if $¬(∃x) P(x)$ holds, then $(∃x)P(x) to R$ holds, for $R$ whatever.
Thus : $(∃x)P(x) to (∀y)Q(y)$ holds.
answered Jan 30 at 16:34
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
$endgroup$
$begingroup$
One question: You say we have to work with prenex normal form but your first step switches to a non-prenex normal form if I'm not wrong?
$endgroup$
– Ayoub Rossi
Jan 30 at 17:04
$begingroup$
Second question: If I recognize a prenex normal form, should I always switch to an equivalent in non-prenex form
$endgroup$
– Ayoub Rossi
Jan 30 at 17:06
add a comment |
$begingroup$
Hint
We have to work with Prenex normal form equivalences.
The first formula is equivalent to :
$((∃x)P(x) to (∀y)Q(y)) ∧ (∃z)¬Q(z)$.
By De Morgan, the second formula is :
$¬(∃x) P(x) ∧ ¬(∀z)Q(z)$, i.e. $¬(∃x) P(x) ∧ (∃z)¬Q(z)$.
The first one implies the second : if we have that $(∃z)¬Q(z)$ holds, then it is false that $(∀y)Q(y)$ and thus also $(∃x)P(x)$ is false.
Thus : $¬(∃x) P(x)$ holds.
The second one implies the first : if $¬(∃x) P(x)$ holds, then $(∃x)P(x) to R$ holds, for $R$ whatever.
Thus : $(∃x)P(x) to (∀y)Q(y)$ holds.
answered Jan 30 at 16:34
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
$endgroup$
$begingroup$
One question: You say we have to work with prenex normal form but your first step switches to a non-prenex normal form if I'm not wrong?
$endgroup$
– Ayoub Rossi
Jan 30 at 17:04
$begingroup$
Second question: If I recognize a prenex normal form, should I always switch to an equivalent in non-prenex form
$endgroup$
– Ayoub Rossi
Jan 30 at 17:06
add a comment |
$begingroup$
Hint
We have to work with Prenex normal form equivalences.
The first formula is equivalent to :
$((∃x)P(x) to (∀y)Q(y)) ∧ (∃z)¬Q(z)$.
By De Morgan, the second formula is :
$¬(∃x) P(x) ∧ ¬(∀z)Q(z)$, i.e. $¬(∃x) P(x) ∧ (∃z)¬Q(z)$.
The first one implies the second : if we have that $(∃z)¬Q(z)$ holds, then it is false that $(∀y)Q(y)$ and thus also $(∃x)P(x)$ is false.
Thus : $¬(∃x) P(x)$ holds.
The second one implies the first : if $¬(∃x) P(x)$ holds, then $(∃x)P(x) to R$ holds, for $R$ whatever.
Thus : $(∃x)P(x) to (∀y)Q(y)$ holds.
answered Jan 30 at 16:34
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
$endgroup$
Hint
We have to work with Prenex normal form equivalences.
The first formula is equivalent to :
$((∃x)P(x) to (∀y)Q(y)) ∧ (∃z)¬Q(z)$.
By De Morgan, the second formula is :
$¬(∃x) P(x) ∧ ¬(∀z)Q(z)$, i.e. $¬(∃x) P(x) ∧ (∃z)¬Q(z)$.
The first one implies the second : if we have that $(∃z)¬Q(z)$ holds, then it is false that $(∀y)Q(y)$ and thus also $(∃x)P(x)$ is false.
Thus : $¬(∃x) P(x)$ holds.
The second one implies the first : if $¬(∃x) P(x)$ holds, then $(∃x)P(x) to R$ holds, for $R$ whatever.
Thus : $(∃x)P(x) to (∀y)Q(y)$ holds.
answered Jan 30 at 16:34
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
answered Jan 30 at 16:34
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
answered Jan 30 at 16:34
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
answered Jan 30 at 16:34
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
67.7k449117
$begingroup$
One question: You say we have to work with prenex normal form but your first step switches to a non-prenex normal form if I'm not wrong?
$endgroup$
– Ayoub Rossi
Jan 30 at 17:04
$begingroup$
Second question: If I recognize a prenex normal form, should I always switch to an equivalent in non-prenex form
$endgroup$
– Ayoub Rossi
Jan 30 at 17:06
add a comment |
$begingroup$
One question: You say we have to work with prenex normal form but your first step switches to a non-prenex normal form if I'm not wrong?
$endgroup$
– Ayoub Rossi
Jan 30 at 17:04
$begingroup$
Second question: If I recognize a prenex normal form, should I always switch to an equivalent in non-prenex form
$endgroup$
– Ayoub Rossi
Jan 30 at 17:06
$begingroup$
One question: You say we have to work with prenex normal form but your first step switches to a non-prenex normal form if I'm not wrong?
$endgroup$
– Ayoub Rossi
Jan 30 at 17:04
$begingroup$
One question: You say we have to work with prenex normal form but your first step switches to a non-prenex normal form if I'm not wrong?
$endgroup$
– Ayoub Rossi
Jan 30 at 17:04
$begingroup$
Second question: If I recognize a prenex normal form, should I always switch to an equivalent in non-prenex form
$endgroup$
– Ayoub Rossi
Jan 30 at 17:06
$begingroup$
Second question: If I recognize a prenex normal form, should I always switch to an equivalent in non-prenex form
$endgroup$
– Ayoub Rossi
Jan 30 at 17:06
add a comment |
$begingroup$
$ forall x forall y exists z((P(x)rightarrow Q(y)) land neg Q(z)) overset{Prenex}{Leftrightarrow}$
$ forall x forall y ((P(x)rightarrow Q(y)) land exists z neg Q(z)) overset{Implication}{Leftrightarrow}$
$ forall x forall y ((neg P(x) lor Q(y)) land exists z neg Q(z))) overset{Distribution}{Leftrightarrow}$
$ forall x forall y ((neg P(x) land exists z neg Q(z)) lor ((Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ forall x (neg P(x) land exists z neg Q(z)) lor forall y (Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ ( forall x neg P(x) land exists z neg Q(z)) lor (forall y Q(y) land exists z neg Q(z)) overset{Quantifier Negation}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall z Q(z)) overset{Replacing Variables}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall y Q(y)) overset{Complement}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor bot overset{Identity}{Leftrightarrow}$
$ neg exists x P(x) land neg forall z Q(z)overset{DeMorgan}{Leftrightarrow}$
$ neg( exists x P(x) lor neg forall z Q(z))$
answered Jan 30 at 22:39
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
$begingroup$
$ forall x forall y exists z((P(x)rightarrow Q(y)) land neg Q(z)) overset{Prenex}{Leftrightarrow}$
$ forall x forall y ((P(x)rightarrow Q(y)) land exists z neg Q(z)) overset{Implication}{Leftrightarrow}$
$ forall x forall y ((neg P(x) lor Q(y)) land exists z neg Q(z))) overset{Distribution}{Leftrightarrow}$
$ forall x forall y ((neg P(x) land exists z neg Q(z)) lor ((Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ forall x (neg P(x) land exists z neg Q(z)) lor forall y (Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ ( forall x neg P(x) land exists z neg Q(z)) lor (forall y Q(y) land exists z neg Q(z)) overset{Quantifier Negation}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall z Q(z)) overset{Replacing Variables}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall y Q(y)) overset{Complement}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor bot overset{Identity}{Leftrightarrow}$
$ neg exists x P(x) land neg forall z Q(z)overset{DeMorgan}{Leftrightarrow}$
$ neg( exists x P(x) lor neg forall z Q(z))$
answered Jan 30 at 22:39
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
$begingroup$
$ forall x forall y exists z((P(x)rightarrow Q(y)) land neg Q(z)) overset{Prenex}{Leftrightarrow}$
$ forall x forall y ((P(x)rightarrow Q(y)) land exists z neg Q(z)) overset{Implication}{Leftrightarrow}$
$ forall x forall y ((neg P(x) lor Q(y)) land exists z neg Q(z))) overset{Distribution}{Leftrightarrow}$
$ forall x forall y ((neg P(x) land exists z neg Q(z)) lor ((Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ forall x (neg P(x) land exists z neg Q(z)) lor forall y (Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ ( forall x neg P(x) land exists z neg Q(z)) lor (forall y Q(y) land exists z neg Q(z)) overset{Quantifier Negation}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall z Q(z)) overset{Replacing Variables}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall y Q(y)) overset{Complement}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor bot overset{Identity}{Leftrightarrow}$
$ neg exists x P(x) land neg forall z Q(z)overset{DeMorgan}{Leftrightarrow}$
$ neg( exists x P(x) lor neg forall z Q(z))$
answered Jan 30 at 22:39
Bram28Bram28
64.2k44793
$endgroup$
$ forall x forall y exists z((P(x)rightarrow Q(y)) land neg Q(z)) overset{Prenex}{Leftrightarrow}$
$ forall x forall y ((P(x)rightarrow Q(y)) land exists z neg Q(z)) overset{Implication}{Leftrightarrow}$
$ forall x forall y ((neg P(x) lor Q(y)) land exists z neg Q(z))) overset{Distribution}{Leftrightarrow}$
$ forall x forall y ((neg P(x) land exists z neg Q(z)) lor ((Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ forall x (neg P(x) land exists z neg Q(z)) lor forall y (Q(y) land exists z neg Q(z)) overset{Prenex x 2}{Leftrightarrow}$
$ ( forall x neg P(x) land exists z neg Q(z)) lor (forall y Q(y) land exists z neg Q(z)) overset{Quantifier Negation}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall z Q(z)) overset{Replacing Variables}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor (forall y Q(y) land neg forall y Q(y)) overset{Complement}{Leftrightarrow}$
$ (neg exists x P(x) land neg forall z Q(z)) lor bot overset{Identity}{Leftrightarrow}$
$ neg exists x P(x) land neg forall z Q(z)overset{DeMorgan}{Leftrightarrow}$
$ neg( exists x P(x) lor neg forall z Q(z))$
answered Jan 30 at 22:39
Bram28Bram28
64.2k44793
edited Jan 31 at 0:37
answered Jan 30 at 22:39
Bram28Bram28
64.2k44793
answered Jan 30 at 22:39
Bram28Bram28
64.2k44793
answered Jan 30 at 22:39
Bram28Bram28
64.2k44793
64.2k44793
add a comment |
add a comment |
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$begingroup$
Is there an $y$ missing in the first formula ?
$endgroup$
– Mauro ALLEGRANZA
Jan 30 at 13:51
$begingroup$
Indeed, thanks!
$endgroup$
– Ayoub Rossi
Jan 30 at 13:54
$begingroup$
Well in the case that $Q(y)$ for all $y$, then it is true for any specific choice of $y$, namely $z$. Hence we can replace $forall y(P(x) rightarrow Q(y))$ with $ (P(x) rightarrow Q(z)) $
$endgroup$
– NazimJ
Jan 30 at 14:37