Generating truth tables for quantifiers online
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Is there any online tool that can generate truth tables for quatifiers (existential and universal). I can generate for Boolean equations not involving quantifier as this one? But I didnt find any example for quantifiers here and here.
Also can we specify more than one equations in wolframalpha, so that it can display truth values for more than one equations side by side in the same truth table as shown below? And also possibly show the truth value for each intermediate operator also shown below?
logic propositional-calculus first-order-logic
asked Dec 27 '15 at 12:22
anir123
855923
|
show 2 more comments
up vote
0
down vote
favorite
Is there any online tool that can generate truth tables for quatifiers (existential and universal). I can generate for Boolean equations not involving quantifier as this one? But I didnt find any example for quantifiers here and here.
Also can we specify more than one equations in wolframalpha, so that it can display truth values for more than one equations side by side in the same truth table as shown below? And also possibly show the truth value for each intermediate operator also shown below?
logic propositional-calculus first-order-logic
asked Dec 27 '15 at 12:22
anir123
855923
You may express all quantified statements in terms of more "elementary" logic and some set theory, for example $forall xin A: P(x)$ is equivalent to $xin Aimplies P(x)$.
– Oskar Limka
Dec 27 '15 at 12:27
@OskarLimka $forall xin A:P(x)$ is equivalent to $forall x,(xin Aimplies P(x))$. You probably intended $forall x$ to be tacitly understood as part of the meaning of $implies$. If so, that should be made clear to the OP since (s)he is likely to interpret $implies$ by just a truth table.
– Andreas Blass
Nov 23 at 18:26
@Andreas Blass quantifying $x$ in the second statement is ok (for emphasis), but it is not logically necessary. The $x$ in the statement $[xin ARightarrow P(x)]$ is a dummy variable (so it doesn't need quantification) whereas in $[P(x)]$ it is not a dummy variable and we need to quantify it. I bracketed propositions. Think of the following example $P(x):Leftrightarrow [xgeq0]$ and $A:=mathbb N$.
– Oskar Limka
Nov 24 at 20:45
@OskarLimka Thanks for confirming my conjecture that you intended $forall x$to be implicit in your formula (since you say $x$ is a dummy variable). I still think it's good to make that clear to the OP, since the usual definitions would say that $x$ is free in $xin Aimplies P(x)$. But I think our comments have, collectively, made that clear enough.
– Andreas Blass
Nov 25 at 6:02
@Andreas Blass Indeed. My rule of thumb redundancy-test (in logic) is to remove something and see if what remains still makes sense. If we remove the quantified bit in $forall xin A:P(x)$, we're left with $P(x)$ which cannot be assigned a true/false value because $x$ is "dangling" (to borrow terms from computer science), so strictly speaking $P(x)$ is not a proposition but a indexed family of propositions (logic folks call a construction such as $P(x)$ "formula"). Whereas in $forall x : xin Aimplies P(x)$ removing the quantifier leaves us with a proposition.
– Oskar Limka
Nov 29 at 8:55
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is there any online tool that can generate truth tables for quatifiers (existential and universal). I can generate for Boolean equations not involving quantifier as this one? But I didnt find any example for quantifiers here and here.
Also can we specify more than one equations in wolframalpha, so that it can display truth values for more than one equations side by side in the same truth table as shown below? And also possibly show the truth value for each intermediate operator also shown below?
logic propositional-calculus first-order-logic
asked Dec 27 '15 at 12:22
anir123
855923
Is there any online tool that can generate truth tables for quatifiers (existential and universal). I can generate for Boolean equations not involving quantifier as this one? But I didnt find any example for quantifiers here and here.
Also can we specify more than one equations in wolframalpha, so that it can display truth values for more than one equations side by side in the same truth table as shown below? And also possibly show the truth value for each intermediate operator also shown below?
logic propositional-calculus first-order-logic
logic propositional-calculus first-order-logic
asked Dec 27 '15 at 12:22
anir123
855923
asked Dec 27 '15 at 12:22
anir123
855923
edited Nov 25 at 9:59
asked Dec 27 '15 at 12:22
anir123
855923
asked Dec 27 '15 at 12:22
anir123
855923
asked Dec 27 '15 at 12:22
anir123
855923
855923
You may express all quantified statements in terms of more "elementary" logic and some set theory, for example $forall xin A: P(x)$ is equivalent to $xin Aimplies P(x)$.
– Oskar Limka
Dec 27 '15 at 12:27
@OskarLimka $forall xin A:P(x)$ is equivalent to $forall x,(xin Aimplies P(x))$. You probably intended $forall x$ to be tacitly understood as part of the meaning of $implies$. If so, that should be made clear to the OP since (s)he is likely to interpret $implies$ by just a truth table.
– Andreas Blass
Nov 23 at 18:26
@Andreas Blass quantifying $x$ in the second statement is ok (for emphasis), but it is not logically necessary. The $x$ in the statement $[xin ARightarrow P(x)]$ is a dummy variable (so it doesn't need quantification) whereas in $[P(x)]$ it is not a dummy variable and we need to quantify it. I bracketed propositions. Think of the following example $P(x):Leftrightarrow [xgeq0]$ and $A:=mathbb N$.
– Oskar Limka
Nov 24 at 20:45
@OskarLimka Thanks for confirming my conjecture that you intended $forall x$to be implicit in your formula (since you say $x$ is a dummy variable). I still think it's good to make that clear to the OP, since the usual definitions would say that $x$ is free in $xin Aimplies P(x)$. But I think our comments have, collectively, made that clear enough.
– Andreas Blass
Nov 25 at 6:02
@Andreas Blass Indeed. My rule of thumb redundancy-test (in logic) is to remove something and see if what remains still makes sense. If we remove the quantified bit in $forall xin A:P(x)$, we're left with $P(x)$ which cannot be assigned a true/false value because $x$ is "dangling" (to borrow terms from computer science), so strictly speaking $P(x)$ is not a proposition but a indexed family of propositions (logic folks call a construction such as $P(x)$ "formula"). Whereas in $forall x : xin Aimplies P(x)$ removing the quantifier leaves us with a proposition.
– Oskar Limka
Nov 29 at 8:55
|
show 2 more comments
You may express all quantified statements in terms of more "elementary" logic and some set theory, for example $forall xin A: P(x)$ is equivalent to $xin Aimplies P(x)$.
– Oskar Limka
Dec 27 '15 at 12:27
@OskarLimka $forall xin A:P(x)$ is equivalent to $forall x,(xin Aimplies P(x))$. You probably intended $forall x$ to be tacitly understood as part of the meaning of $implies$. If so, that should be made clear to the OP since (s)he is likely to interpret $implies$ by just a truth table.
– Andreas Blass
Nov 23 at 18:26
@Andreas Blass quantifying $x$ in the second statement is ok (for emphasis), but it is not logically necessary. The $x$ in the statement $[xin ARightarrow P(x)]$ is a dummy variable (so it doesn't need quantification) whereas in $[P(x)]$ it is not a dummy variable and we need to quantify it. I bracketed propositions. Think of the following example $P(x):Leftrightarrow [xgeq0]$ and $A:=mathbb N$.
– Oskar Limka
Nov 24 at 20:45
@OskarLimka Thanks for confirming my conjecture that you intended $forall x$to be implicit in your formula (since you say $x$ is a dummy variable). I still think it's good to make that clear to the OP, since the usual definitions would say that $x$ is free in $xin Aimplies P(x)$. But I think our comments have, collectively, made that clear enough.
– Andreas Blass
Nov 25 at 6:02
@Andreas Blass Indeed. My rule of thumb redundancy-test (in logic) is to remove something and see if what remains still makes sense. If we remove the quantified bit in $forall xin A:P(x)$, we're left with $P(x)$ which cannot be assigned a true/false value because $x$ is "dangling" (to borrow terms from computer science), so strictly speaking $P(x)$ is not a proposition but a indexed family of propositions (logic folks call a construction such as $P(x)$ "formula"). Whereas in $forall x : xin Aimplies P(x)$ removing the quantifier leaves us with a proposition.
– Oskar Limka
Nov 29 at 8:55
You may express all quantified statements in terms of more "elementary" logic and some set theory, for example $forall xin A: P(x)$ is equivalent to $xin Aimplies P(x)$.
– Oskar Limka
Dec 27 '15 at 12:27
You may express all quantified statements in terms of more "elementary" logic and some set theory, for example $forall xin A: P(x)$ is equivalent to $xin Aimplies P(x)$.
– Oskar Limka
Dec 27 '15 at 12:27
@OskarLimka $forall xin A:P(x)$ is equivalent to $forall x,(xin Aimplies P(x))$. You probably intended $forall x$ to be tacitly understood as part of the meaning of $implies$. If so, that should be made clear to the OP since (s)he is likely to interpret $implies$ by just a truth table.
– Andreas Blass
Nov 23 at 18:26
@OskarLimka $forall xin A:P(x)$ is equivalent to $forall x,(xin Aimplies P(x))$. You probably intended $forall x$ to be tacitly understood as part of the meaning of $implies$. If so, that should be made clear to the OP since (s)he is likely to interpret $implies$ by just a truth table.
– Andreas Blass
Nov 23 at 18:26
@Andreas Blass quantifying $x$ in the second statement is ok (for emphasis), but it is not logically necessary. The $x$ in the statement $[xin ARightarrow P(x)]$ is a dummy variable (so it doesn't need quantification) whereas in $[P(x)]$ it is not a dummy variable and we need to quantify it. I bracketed propositions. Think of the following example $P(x):Leftrightarrow [xgeq0]$ and $A:=mathbb N$.
– Oskar Limka
Nov 24 at 20:45
@Andreas Blass quantifying $x$ in the second statement is ok (for emphasis), but it is not logically necessary. The $x$ in the statement $[xin ARightarrow P(x)]$ is a dummy variable (so it doesn't need quantification) whereas in $[P(x)]$ it is not a dummy variable and we need to quantify it. I bracketed propositions. Think of the following example $P(x):Leftrightarrow [xgeq0]$ and $A:=mathbb N$.
– Oskar Limka
Nov 24 at 20:45
@OskarLimka Thanks for confirming my conjecture that you intended $forall x$to be implicit in your formula (since you say $x$ is a dummy variable). I still think it's good to make that clear to the OP, since the usual definitions would say that $x$ is free in $xin Aimplies P(x)$. But I think our comments have, collectively, made that clear enough.
– Andreas Blass
Nov 25 at 6:02
@OskarLimka Thanks for confirming my conjecture that you intended $forall x$to be implicit in your formula (since you say $x$ is a dummy variable). I still think it's good to make that clear to the OP, since the usual definitions would say that $x$ is free in $xin Aimplies P(x)$. But I think our comments have, collectively, made that clear enough.
– Andreas Blass
Nov 25 at 6:02
@Andreas Blass Indeed. My rule of thumb redundancy-test (in logic) is to remove something and see if what remains still makes sense. If we remove the quantified bit in $forall xin A:P(x)$, we're left with $P(x)$ which cannot be assigned a true/false value because $x$ is "dangling" (to borrow terms from computer science), so strictly speaking $P(x)$ is not a proposition but a indexed family of propositions (logic folks call a construction such as $P(x)$ "formula"). Whereas in $forall x : xin Aimplies P(x)$ removing the quantifier leaves us with a proposition.
– Oskar Limka
Nov 29 at 8:55
@Andreas Blass Indeed. My rule of thumb redundancy-test (in logic) is to remove something and see if what remains still makes sense. If we remove the quantified bit in $forall xin A:P(x)$, we're left with $P(x)$ which cannot be assigned a true/false value because $x$ is "dangling" (to borrow terms from computer science), so strictly speaking $P(x)$ is not a proposition but a indexed family of propositions (logic folks call a construction such as $P(x)$ "formula"). Whereas in $forall x : xin Aimplies P(x)$ removing the quantifier leaves us with a proposition.
– Oskar Limka
Nov 29 at 8:55
|
show 2 more comments
1 Answer
1
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up vote
1
down vote
The semantics for quantifiers are more complicated than truth tables can deal with. If $forall x$ was defined via truth table, you would have to give meaning to the formula $P(x)$, so that $(forall x)P(x)$ can have a truth value. But, $x$ is a variable, so, $P(x)$ isn't a claim that it makes sense to assign a truth value to, without a way of interpreting that variable.
answered Dec 27 '15 at 15:23
James
4,2301821
hmm that makes sense, but then is there any way to check whether the given implication or equivalence involving quantifiers are true? In fact, I came across this problem, which I am able to solve it with the help of verbal interpretation, but not mathematically (can you also help me with solving this mathematically?). I was thinking if I can check them through some tool.
– anir123
Dec 27 '15 at 15:44
There is no procedure which, given a sentence in FOL, will return yes if it is a tautology, and no otherwise. This means that you have to approach in an ad hoc fashion. If you think it is true, try to give a proof, if you think it is false, try to find a counterexample.
– James
Dec 27 '15 at 16:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The semantics for quantifiers are more complicated than truth tables can deal with. If $forall x$ was defined via truth table, you would have to give meaning to the formula $P(x)$, so that $(forall x)P(x)$ can have a truth value. But, $x$ is a variable, so, $P(x)$ isn't a claim that it makes sense to assign a truth value to, without a way of interpreting that variable.
answered Dec 27 '15 at 15:23
James
4,2301821
hmm that makes sense, but then is there any way to check whether the given implication or equivalence involving quantifiers are true? In fact, I came across this problem, which I am able to solve it with the help of verbal interpretation, but not mathematically (can you also help me with solving this mathematically?). I was thinking if I can check them through some tool.
– anir123
Dec 27 '15 at 15:44
There is no procedure which, given a sentence in FOL, will return yes if it is a tautology, and no otherwise. This means that you have to approach in an ad hoc fashion. If you think it is true, try to give a proof, if you think it is false, try to find a counterexample.
– James
Dec 27 '15 at 16:30
add a comment |
up vote
1
down vote
The semantics for quantifiers are more complicated than truth tables can deal with. If $forall x$ was defined via truth table, you would have to give meaning to the formula $P(x)$, so that $(forall x)P(x)$ can have a truth value. But, $x$ is a variable, so, $P(x)$ isn't a claim that it makes sense to assign a truth value to, without a way of interpreting that variable.
answered Dec 27 '15 at 15:23
James
4,2301821
hmm that makes sense, but then is there any way to check whether the given implication or equivalence involving quantifiers are true? In fact, I came across this problem, which I am able to solve it with the help of verbal interpretation, but not mathematically (can you also help me with solving this mathematically?). I was thinking if I can check them through some tool.
– anir123
Dec 27 '15 at 15:44
There is no procedure which, given a sentence in FOL, will return yes if it is a tautology, and no otherwise. This means that you have to approach in an ad hoc fashion. If you think it is true, try to give a proof, if you think it is false, try to find a counterexample.
– James
Dec 27 '15 at 16:30
add a comment |
up vote
1
down vote
up vote
1
down vote
The semantics for quantifiers are more complicated than truth tables can deal with. If $forall x$ was defined via truth table, you would have to give meaning to the formula $P(x)$, so that $(forall x)P(x)$ can have a truth value. But, $x$ is a variable, so, $P(x)$ isn't a claim that it makes sense to assign a truth value to, without a way of interpreting that variable.
answered Dec 27 '15 at 15:23
James
4,2301821
The semantics for quantifiers are more complicated than truth tables can deal with. If $forall x$ was defined via truth table, you would have to give meaning to the formula $P(x)$, so that $(forall x)P(x)$ can have a truth value. But, $x$ is a variable, so, $P(x)$ isn't a claim that it makes sense to assign a truth value to, without a way of interpreting that variable.
answered Dec 27 '15 at 15:23
James
4,2301821
answered Dec 27 '15 at 15:23
James
4,2301821
answered Dec 27 '15 at 15:23
James
4,2301821
answered Dec 27 '15 at 15:23
James
4,2301821
4,2301821
hmm that makes sense, but then is there any way to check whether the given implication or equivalence involving quantifiers are true? In fact, I came across this problem, which I am able to solve it with the help of verbal interpretation, but not mathematically (can you also help me with solving this mathematically?). I was thinking if I can check them through some tool.
– anir123
Dec 27 '15 at 15:44
There is no procedure which, given a sentence in FOL, will return yes if it is a tautology, and no otherwise. This means that you have to approach in an ad hoc fashion. If you think it is true, try to give a proof, if you think it is false, try to find a counterexample.
– James
Dec 27 '15 at 16:30
add a comment |
hmm that makes sense, but then is there any way to check whether the given implication or equivalence involving quantifiers are true? In fact, I came across this problem, which I am able to solve it with the help of verbal interpretation, but not mathematically (can you also help me with solving this mathematically?). I was thinking if I can check them through some tool.
– anir123
Dec 27 '15 at 15:44
There is no procedure which, given a sentence in FOL, will return yes if it is a tautology, and no otherwise. This means that you have to approach in an ad hoc fashion. If you think it is true, try to give a proof, if you think it is false, try to find a counterexample.
– James
Dec 27 '15 at 16:30
hmm that makes sense, but then is there any way to check whether the given implication or equivalence involving quantifiers are true? In fact, I came across this problem, which I am able to solve it with the help of verbal interpretation, but not mathematically (can you also help me with solving this mathematically?). I was thinking if I can check them through some tool.
– anir123
Dec 27 '15 at 15:44
hmm that makes sense, but then is there any way to check whether the given implication or equivalence involving quantifiers are true? In fact, I came across this problem, which I am able to solve it with the help of verbal interpretation, but not mathematically (can you also help me with solving this mathematically?). I was thinking if I can check them through some tool.
– anir123
Dec 27 '15 at 15:44
There is no procedure which, given a sentence in FOL, will return yes if it is a tautology, and no otherwise. This means that you have to approach in an ad hoc fashion. If you think it is true, try to give a proof, if you think it is false, try to find a counterexample.
– James
Dec 27 '15 at 16:30
There is no procedure which, given a sentence in FOL, will return yes if it is a tautology, and no otherwise. This means that you have to approach in an ad hoc fashion. If you think it is true, try to give a proof, if you think it is false, try to find a counterexample.
– James
Dec 27 '15 at 16:30
add a comment |
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You may express all quantified statements in terms of more "elementary" logic and some set theory, for example $forall xin A: P(x)$ is equivalent to $xin Aimplies P(x)$.
– Oskar Limka
Dec 27 '15 at 12:27
@OskarLimka $forall xin A:P(x)$ is equivalent to $forall x,(xin Aimplies P(x))$. You probably intended $forall x$ to be tacitly understood as part of the meaning of $implies$. If so, that should be made clear to the OP since (s)he is likely to interpret $implies$ by just a truth table.
– Andreas Blass
Nov 23 at 18:26
@Andreas Blass quantifying $x$ in the second statement is ok (for emphasis), but it is not logically necessary. The $x$ in the statement $[xin ARightarrow P(x)]$ is a dummy variable (so it doesn't need quantification) whereas in $[P(x)]$ it is not a dummy variable and we need to quantify it. I bracketed propositions. Think of the following example $P(x):Leftrightarrow [xgeq0]$ and $A:=mathbb N$.
– Oskar Limka
Nov 24 at 20:45
@OskarLimka Thanks for confirming my conjecture that you intended $forall x$to be implicit in your formula (since you say $x$ is a dummy variable). I still think it's good to make that clear to the OP, since the usual definitions would say that $x$ is free in $xin Aimplies P(x)$. But I think our comments have, collectively, made that clear enough.
– Andreas Blass
Nov 25 at 6:02
@Andreas Blass Indeed. My rule of thumb redundancy-test (in logic) is to remove something and see if what remains still makes sense. If we remove the quantified bit in $forall xin A:P(x)$, we're left with $P(x)$ which cannot be assigned a true/false value because $x$ is "dangling" (to borrow terms from computer science), so strictly speaking $P(x)$ is not a proposition but a indexed family of propositions (logic folks call a construction such as $P(x)$ "formula"). Whereas in $forall x : xin Aimplies P(x)$ removing the quantifier leaves us with a proposition.
– Oskar Limka
Nov 29 at 8:55