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How the definition of the truth table of the material implication (&p⇒q&), is “derived†from the definition of a valid argument?
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According to my book
An argument is valid if and only if it's impossible that the premises are true and the conclusion false.
So if I have the argument $p_1∧p_2∧...∧p_t⇒q$ and I want to prove that it's valid, I suppose that all premises are true and the conclusion is false, and if I get to a contradiction, then I have proven that it's impossible for the premises to be true and the conclusion false, hence making it a valid argument.
What I want to understand is how the definition of the truth table of the material implication "models"/"agrees"/"is derived", etc from this definition.
When I want to prove if $p_1∧p_2∧...∧p_t⇒q$ is valid, I assume that all the premises are true and the conclusion is false, right? like this : $V_1∧V_2∧...∧V_t⇒F$, which makes the conditional false (according to the third line of the pic of the truth table).
So, even though I have never understood why the conditional is defined the way it's, from the latter I conjecture this:
Again the definition of a valid argument is
An argument is valid if and only if it's impossible that the premises are true and the conclusion false.
It doesn't say anything about arguments with true premises and true conclusions, or arguments with false premises and true conclusions, or arguments with false premises and false conclusions. I assume that this are valid arguments. coincidentally these are also the ones which are true in the truth table.
The definition only refers to the case in which the premises are true and the conclusion false, as an invalid argument. coincidentally the only one which is false in the truth table.
Now, my teacher says that an argument is neither true nor false, but valid or invalid, while the ones that are either true or false are propositions; However, it seems to me, from what I have said, that there is a certain relation between the falsehood or veracity of a material conditional, and the validity of the argument it models; however this doesn't completely "click" in my brain, and I don't know if there are any mistakes in my reasoning.
I also want to know how this relates with the definition of a valid argument in propositional logic.
$p_1∧p_2∧...∧p_t⇒q$ is valid if and only if $p_1∧p_2∧...∧p_t⇒q$ is a tautology.
I'm pretty sure it follows from all the aforementioned, but it still doesn't completely "click" in my brain.
Thanks in advance.
logic propositional-calculus
asked Jul 21 at 21:21
Daniel Bonilla Jaramillo
38819
add a comment |Â
up vote
3
down vote
favorite
According to my book
An argument is valid if and only if it's impossible that the premises are true and the conclusion false.
So if I have the argument $p_1∧p_2∧...∧p_t⇒q$ and I want to prove that it's valid, I suppose that all premises are true and the conclusion is false, and if I get to a contradiction, then I have proven that it's impossible for the premises to be true and the conclusion false, hence making it a valid argument.
What I want to understand is how the definition of the truth table of the material implication "models"/"agrees"/"is derived", etc from this definition.
When I want to prove if $p_1∧p_2∧...∧p_t⇒q$ is valid, I assume that all the premises are true and the conclusion is false, right? like this : $V_1∧V_2∧...∧V_t⇒F$, which makes the conditional false (according to the third line of the pic of the truth table).
So, even though I have never understood why the conditional is defined the way it's, from the latter I conjecture this:
Again the definition of a valid argument is
An argument is valid if and only if it's impossible that the premises are true and the conclusion false.
It doesn't say anything about arguments with true premises and true conclusions, or arguments with false premises and true conclusions, or arguments with false premises and false conclusions. I assume that this are valid arguments. coincidentally these are also the ones which are true in the truth table.
The definition only refers to the case in which the premises are true and the conclusion false, as an invalid argument. coincidentally the only one which is false in the truth table.
Now, my teacher says that an argument is neither true nor false, but valid or invalid, while the ones that are either true or false are propositions; However, it seems to me, from what I have said, that there is a certain relation between the falsehood or veracity of a material conditional, and the validity of the argument it models; however this doesn't completely "click" in my brain, and I don't know if there are any mistakes in my reasoning.
I also want to know how this relates with the definition of a valid argument in propositional logic.
$p_1∧p_2∧...∧p_t⇒q$ is valid if and only if $p_1∧p_2∧...∧p_t⇒q$ is a tautology.
I'm pretty sure it follows from all the aforementioned, but it still doesn't completely "click" in my brain.
Thanks in advance.
logic propositional-calculus
asked Jul 21 at 21:21
Daniel Bonilla Jaramillo
38819
1
See my formal proof at dcproof.com/DeriveImplies.html There I derive the usual "definition" of material implication using what I believe to be elementary, self-evident rules of inference in natural deduction. Also see my recent blog posting dcproof.com/IfPigsCanFly.html
– Dan Christensen
Jul 22 at 3:51
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
According to my book
An argument is valid if and only if it's impossible that the premises are true and the conclusion false.
So if I have the argument $p_1∧p_2∧...∧p_t⇒q$ and I want to prove that it's valid, I suppose that all premises are true and the conclusion is false, and if I get to a contradiction, then I have proven that it's impossible for the premises to be true and the conclusion false, hence making it a valid argument.
What I want to understand is how the definition of the truth table of the material implication "models"/"agrees"/"is derived", etc from this definition.
When I want to prove if $p_1∧p_2∧...∧p_t⇒q$ is valid, I assume that all the premises are true and the conclusion is false, right? like this : $V_1∧V_2∧...∧V_t⇒F$, which makes the conditional false (according to the third line of the pic of the truth table).
So, even though I have never understood why the conditional is defined the way it's, from the latter I conjecture this:
Again the definition of a valid argument is
An argument is valid if and only if it's impossible that the premises are true and the conclusion false.
It doesn't say anything about arguments with true premises and true conclusions, or arguments with false premises and true conclusions, or arguments with false premises and false conclusions. I assume that this are valid arguments. coincidentally these are also the ones which are true in the truth table.
The definition only refers to the case in which the premises are true and the conclusion false, as an invalid argument. coincidentally the only one which is false in the truth table.
Now, my teacher says that an argument is neither true nor false, but valid or invalid, while the ones that are either true or false are propositions; However, it seems to me, from what I have said, that there is a certain relation between the falsehood or veracity of a material conditional, and the validity of the argument it models; however this doesn't completely "click" in my brain, and I don't know if there are any mistakes in my reasoning.
I also want to know how this relates with the definition of a valid argument in propositional logic.
$p_1∧p_2∧...∧p_t⇒q$ is valid if and only if $p_1∧p_2∧...∧p_t⇒q$ is a tautology.
I'm pretty sure it follows from all the aforementioned, but it still doesn't completely "click" in my brain.
Thanks in advance.
logic propositional-calculus
asked Jul 21 at 21:21
Daniel Bonilla Jaramillo
38819
According to my book
An argument is valid if and only if it's impossible that the premises are true and the conclusion false.
So if I have the argument $p_1∧p_2∧...∧p_t⇒q$ and I want to prove that it's valid, I suppose that all premises are true and the conclusion is false, and if I get to a contradiction, then I have proven that it's impossible for the premises to be true and the conclusion false, hence making it a valid argument.
What I want to understand is how the definition of the truth table of the material implication "models"/"agrees"/"is derived", etc from this definition.
When I want to prove if $p_1∧p_2∧...∧p_t⇒q$ is valid, I assume that all the premises are true and the conclusion is false, right? like this : $V_1∧V_2∧...∧V_t⇒F$, which makes the conditional false (according to the third line of the pic of the truth table).
So, even though I have never understood why the conditional is defined the way it's, from the latter I conjecture this:
Again the definition of a valid argument is
An argument is valid if and only if it's impossible that the premises are true and the conclusion false.
It doesn't say anything about arguments with true premises and true conclusions, or arguments with false premises and true conclusions, or arguments with false premises and false conclusions. I assume that this are valid arguments. coincidentally these are also the ones which are true in the truth table.
The definition only refers to the case in which the premises are true and the conclusion false, as an invalid argument. coincidentally the only one which is false in the truth table.
Now, my teacher says that an argument is neither true nor false, but valid or invalid, while the ones that are either true or false are propositions; However, it seems to me, from what I have said, that there is a certain relation between the falsehood or veracity of a material conditional, and the validity of the argument it models; however this doesn't completely "click" in my brain, and I don't know if there are any mistakes in my reasoning.
I also want to know how this relates with the definition of a valid argument in propositional logic.
$p_1∧p_2∧...∧p_t⇒q$ is valid if and only if $p_1∧p_2∧...∧p_t⇒q$ is a tautology.
I'm pretty sure it follows from all the aforementioned, but it still doesn't completely "click" in my brain.
Thanks in advance.
logic propositional-calculus
asked Jul 21 at 21:21
Daniel Bonilla Jaramillo
38819
asked Jul 21 at 21:21
Daniel Bonilla Jaramillo
38819
asked Jul 21 at 21:21
Daniel Bonilla Jaramillo
38819
asked Jul 21 at 21:21
Daniel Bonilla Jaramillo
38819
38819
1
See my formal proof at dcproof.com/DeriveImplies.html There I derive the usual "definition" of material implication using what I believe to be elementary, self-evident rules of inference in natural deduction. Also see my recent blog posting dcproof.com/IfPigsCanFly.html
– Dan Christensen
Jul 22 at 3:51
add a comment |Â
1
See my formal proof at dcproof.com/DeriveImplies.html There I derive the usual "definition" of material implication using what I believe to be elementary, self-evident rules of inference in natural deduction. Also see my recent blog posting dcproof.com/IfPigsCanFly.html
– Dan Christensen
Jul 22 at 3:51
1
1
See my formal proof at dcproof.com/DeriveImplies.html There I derive the usual "definition" of material implication using what I believe to be elementary, self-evident rules of inference in natural deduction. Also see my recent blog posting dcproof.com/IfPigsCanFly.html
– Dan Christensen
Jul 22 at 3:51
See my formal proof at dcproof.com/DeriveImplies.html There I derive the usual "definition" of material implication using what I believe to be elementary, self-evident rules of inference in natural deduction. Also see my recent blog posting dcproof.com/IfPigsCanFly.html
– Dan Christensen
Jul 22 at 3:51
add a comment |Â
2 Answers
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First off, if we have an argument with multiple premises, then we don't have a conjunction assumed. If we had a conjunction assumed, we would have only one premise instead of many premises. So, instead of having p1∧p2∧...∧pt⇒q, we have p1,p2,...,pt⇒q, where each pn is a member of the set of premises. Additionally, some propositional calculi don't even have a conjunction connective.
Suppose that it is impossible that all of the premises are true, and the conclusion false. Suppose also that there exists but one premise p, and a conclusion q. By the definition you've given, the argument p⇒q is valid. There's a meta-theorem of propositional logic, called the deduction theorem, which says that if p⇒q, then ⇒(p$rightarrow$q), where $rightarrow$ corresponds to your horseshoe symbol. ⇒(p$rightarrow$q) is another way of saying that (p$rightarrow$q) is true. The only case excluded was when p was true and q false. So for any valid argument with premise p and conclusion q, it follows that if p is false, and q is false, (p$rightarrow$q) is true. If p is false, and q is true, then (p$rightarrow$q) is true. If p is true, and q is true, then (p$rightarrow$q) is true. And there's the definition of the truth table, given that propositions can only take on two truth values.
If we don't have "if p⇒q, then ⇒(p$rightarrow$q)", nor have something like a truth table for $rightarrow$, then it's probably not clear what $rightarrow$ means in the first place. We can't define $rightarrow$ merely by modus ponens, since logical equivalence also has a rule of modus ponens formally speaking.
answered Jul 21 at 23:21
Doug Spoonwood
7,67512042
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up vote
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A fundamental distinction here is that an argument is often about many similar situations that the argument treats uniformly. Often this can be seen because some variable appears both in the premise and the conclusion:
Assume that $x$ is an even number.
Then $x^2$ is an even number.
Here we can see the $x$ on both sides, and we can apply your definition directly: The argument is valid because it is impossible that $x$ is even and $x^2$ is not even.
(Sometimes the shared variable is hidden in natural language: "If a number is even, then its square is even too". That makes no formal difference).
In contrast the connective $supset$ and its truth table is in principle always something you apply to a single situations. The formula
$$ xtext is even supset x^2text is even $$
does not have a truth value until you plug in a concrete value for $x$. Once you do that, you can start applying the truth table:
$$ 4text is even supset 16text is even $$
is true due to the first line of the truth table, and
$$ 5text is even supset 25text is even $$
is true due to the last line of the truth table.
So what we can say is that
- The argument "If $p(x)$ then $q(x)$" is valid, and
- The formula "$p(x)supset q(x)$" is always true, meaning for every value of $x$.
are two ways of making the same claim. With quantifiers we can express the second of these more succinctly as
- The formula "$forall x(p(x)supset q(x)) $" is true. (No "always" here).
The truth table for $supset$ can really only be understood when one knows that it is designed to be used together with an (implicit or explicit) quantification over all the situations one is interested in.
Unfortunately a large majority of introductory texts expect students to understand $supset$ before they even begin to speak about quantifiers, and they spend huge amounts of verbiage trying to cajole the reader into thinking it makes sense even then. It doesn't really.
answered Jul 22 at 0:40
Henning Makholm
226k16290518
If "all men are mortal" and "Henning is a man", then "Henning is mortal" works fine with the conditional. It doesn't require any sort of modern universal or existential quantification. So, I don't know why you've insisted that quantification works as necessary. Also, if you have when p is true, and q is false, and have that (p$rightarrow$p) and (p$rightarrow$(q$rightarrow$p)) are both tautologies, and there exist only two truth values, "true" and "false", then all rows of the truth table follow.
– Doug Spoonwood
Jul 22 at 8:15
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
First off, if we have an argument with multiple premises, then we don't have a conjunction assumed. If we had a conjunction assumed, we would have only one premise instead of many premises. So, instead of having p1∧p2∧...∧pt⇒q, we have p1,p2,...,pt⇒q, where each pn is a member of the set of premises. Additionally, some propositional calculi don't even have a conjunction connective.
Suppose that it is impossible that all of the premises are true, and the conclusion false. Suppose also that there exists but one premise p, and a conclusion q. By the definition you've given, the argument p⇒q is valid. There's a meta-theorem of propositional logic, called the deduction theorem, which says that if p⇒q, then ⇒(p$rightarrow$q), where $rightarrow$ corresponds to your horseshoe symbol. ⇒(p$rightarrow$q) is another way of saying that (p$rightarrow$q) is true. The only case excluded was when p was true and q false. So for any valid argument with premise p and conclusion q, it follows that if p is false, and q is false, (p$rightarrow$q) is true. If p is false, and q is true, then (p$rightarrow$q) is true. If p is true, and q is true, then (p$rightarrow$q) is true. And there's the definition of the truth table, given that propositions can only take on two truth values.
If we don't have "if p⇒q, then ⇒(p$rightarrow$q)", nor have something like a truth table for $rightarrow$, then it's probably not clear what $rightarrow$ means in the first place. We can't define $rightarrow$ merely by modus ponens, since logical equivalence also has a rule of modus ponens formally speaking.
answered Jul 21 at 23:21
Doug Spoonwood
7,67512042
add a comment |Â
up vote
2
down vote
First off, if we have an argument with multiple premises, then we don't have a conjunction assumed. If we had a conjunction assumed, we would have only one premise instead of many premises. So, instead of having p1∧p2∧...∧pt⇒q, we have p1,p2,...,pt⇒q, where each pn is a member of the set of premises. Additionally, some propositional calculi don't even have a conjunction connective.
Suppose that it is impossible that all of the premises are true, and the conclusion false. Suppose also that there exists but one premise p, and a conclusion q. By the definition you've given, the argument p⇒q is valid. There's a meta-theorem of propositional logic, called the deduction theorem, which says that if p⇒q, then ⇒(p$rightarrow$q), where $rightarrow$ corresponds to your horseshoe symbol. ⇒(p$rightarrow$q) is another way of saying that (p$rightarrow$q) is true. The only case excluded was when p was true and q false. So for any valid argument with premise p and conclusion q, it follows that if p is false, and q is false, (p$rightarrow$q) is true. If p is false, and q is true, then (p$rightarrow$q) is true. If p is true, and q is true, then (p$rightarrow$q) is true. And there's the definition of the truth table, given that propositions can only take on two truth values.
If we don't have "if p⇒q, then ⇒(p$rightarrow$q)", nor have something like a truth table for $rightarrow$, then it's probably not clear what $rightarrow$ means in the first place. We can't define $rightarrow$ merely by modus ponens, since logical equivalence also has a rule of modus ponens formally speaking.
answered Jul 21 at 23:21
Doug Spoonwood
7,67512042
add a comment |Â
up vote
2
down vote
up vote
2
down vote
First off, if we have an argument with multiple premises, then we don't have a conjunction assumed. If we had a conjunction assumed, we would have only one premise instead of many premises. So, instead of having p1∧p2∧...∧pt⇒q, we have p1,p2,...,pt⇒q, where each pn is a member of the set of premises. Additionally, some propositional calculi don't even have a conjunction connective.
Suppose that it is impossible that all of the premises are true, and the conclusion false. Suppose also that there exists but one premise p, and a conclusion q. By the definition you've given, the argument p⇒q is valid. There's a meta-theorem of propositional logic, called the deduction theorem, which says that if p⇒q, then ⇒(p$rightarrow$q), where $rightarrow$ corresponds to your horseshoe symbol. ⇒(p$rightarrow$q) is another way of saying that (p$rightarrow$q) is true. The only case excluded was when p was true and q false. So for any valid argument with premise p and conclusion q, it follows that if p is false, and q is false, (p$rightarrow$q) is true. If p is false, and q is true, then (p$rightarrow$q) is true. If p is true, and q is true, then (p$rightarrow$q) is true. And there's the definition of the truth table, given that propositions can only take on two truth values.
If we don't have "if p⇒q, then ⇒(p$rightarrow$q)", nor have something like a truth table for $rightarrow$, then it's probably not clear what $rightarrow$ means in the first place. We can't define $rightarrow$ merely by modus ponens, since logical equivalence also has a rule of modus ponens formally speaking.
answered Jul 21 at 23:21
Doug Spoonwood
7,67512042
First off, if we have an argument with multiple premises, then we don't have a conjunction assumed. If we had a conjunction assumed, we would have only one premise instead of many premises. So, instead of having p1∧p2∧...∧pt⇒q, we have p1,p2,...,pt⇒q, where each pn is a member of the set of premises. Additionally, some propositional calculi don't even have a conjunction connective.
Suppose that it is impossible that all of the premises are true, and the conclusion false. Suppose also that there exists but one premise p, and a conclusion q. By the definition you've given, the argument p⇒q is valid. There's a meta-theorem of propositional logic, called the deduction theorem, which says that if p⇒q, then ⇒(p$rightarrow$q), where $rightarrow$ corresponds to your horseshoe symbol. ⇒(p$rightarrow$q) is another way of saying that (p$rightarrow$q) is true. The only case excluded was when p was true and q false. So for any valid argument with premise p and conclusion q, it follows that if p is false, and q is false, (p$rightarrow$q) is true. If p is false, and q is true, then (p$rightarrow$q) is true. If p is true, and q is true, then (p$rightarrow$q) is true. And there's the definition of the truth table, given that propositions can only take on two truth values.
If we don't have "if p⇒q, then ⇒(p$rightarrow$q)", nor have something like a truth table for $rightarrow$, then it's probably not clear what $rightarrow$ means in the first place. We can't define $rightarrow$ merely by modus ponens, since logical equivalence also has a rule of modus ponens formally speaking.
answered Jul 21 at 23:21
Doug Spoonwood
7,67512042
answered Jul 21 at 23:21
Doug Spoonwood
7,67512042
answered Jul 21 at 23:21
Doug Spoonwood
7,67512042
answered Jul 21 at 23:21
Doug Spoonwood
7,67512042
7,67512042
add a comment |Â
add a comment |Â
up vote
1
down vote
A fundamental distinction here is that an argument is often about many similar situations that the argument treats uniformly. Often this can be seen because some variable appears both in the premise and the conclusion:
Assume that $x$ is an even number.
Then $x^2$ is an even number.
Here we can see the $x$ on both sides, and we can apply your definition directly: The argument is valid because it is impossible that $x$ is even and $x^2$ is not even.
(Sometimes the shared variable is hidden in natural language: "If a number is even, then its square is even too". That makes no formal difference).
In contrast the connective $supset$ and its truth table is in principle always something you apply to a single situations. The formula
$$ xtext is even supset x^2text is even $$
does not have a truth value until you plug in a concrete value for $x$. Once you do that, you can start applying the truth table:
$$ 4text is even supset 16text is even $$
is true due to the first line of the truth table, and
$$ 5text is even supset 25text is even $$
is true due to the last line of the truth table.
So what we can say is that
- The argument "If $p(x)$ then $q(x)$" is valid, and
- The formula "$p(x)supset q(x)$" is always true, meaning for every value of $x$.
are two ways of making the same claim. With quantifiers we can express the second of these more succinctly as
- The formula "$forall x(p(x)supset q(x)) $" is true. (No "always" here).
The truth table for $supset$ can really only be understood when one knows that it is designed to be used together with an (implicit or explicit) quantification over all the situations one is interested in.
Unfortunately a large majority of introductory texts expect students to understand $supset$ before they even begin to speak about quantifiers, and they spend huge amounts of verbiage trying to cajole the reader into thinking it makes sense even then. It doesn't really.
answered Jul 22 at 0:40
Henning Makholm
226k16290518
If "all men are mortal" and "Henning is a man", then "Henning is mortal" works fine with the conditional. It doesn't require any sort of modern universal or existential quantification. So, I don't know why you've insisted that quantification works as necessary. Also, if you have when p is true, and q is false, and have that (p$rightarrow$p) and (p$rightarrow$(q$rightarrow$p)) are both tautologies, and there exist only two truth values, "true" and "false", then all rows of the truth table follow.
– Doug Spoonwood
Jul 22 at 8:15
add a comment |Â
up vote
1
down vote
A fundamental distinction here is that an argument is often about many similar situations that the argument treats uniformly. Often this can be seen because some variable appears both in the premise and the conclusion:
Assume that $x$ is an even number.
Then $x^2$ is an even number.
Here we can see the $x$ on both sides, and we can apply your definition directly: The argument is valid because it is impossible that $x$ is even and $x^2$ is not even.
(Sometimes the shared variable is hidden in natural language: "If a number is even, then its square is even too". That makes no formal difference).
In contrast the connective $supset$ and its truth table is in principle always something you apply to a single situations. The formula
$$ xtext is even supset x^2text is even $$
does not have a truth value until you plug in a concrete value for $x$. Once you do that, you can start applying the truth table:
$$ 4text is even supset 16text is even $$
is true due to the first line of the truth table, and
$$ 5text is even supset 25text is even $$
is true due to the last line of the truth table.
So what we can say is that
- The argument "If $p(x)$ then $q(x)$" is valid, and
- The formula "$p(x)supset q(x)$" is always true, meaning for every value of $x$.
are two ways of making the same claim. With quantifiers we can express the second of these more succinctly as
- The formula "$forall x(p(x)supset q(x)) $" is true. (No "always" here).
The truth table for $supset$ can really only be understood when one knows that it is designed to be used together with an (implicit or explicit) quantification over all the situations one is interested in.
Unfortunately a large majority of introductory texts expect students to understand $supset$ before they even begin to speak about quantifiers, and they spend huge amounts of verbiage trying to cajole the reader into thinking it makes sense even then. It doesn't really.
answered Jul 22 at 0:40
Henning Makholm
226k16290518
If "all men are mortal" and "Henning is a man", then "Henning is mortal" works fine with the conditional. It doesn't require any sort of modern universal or existential quantification. So, I don't know why you've insisted that quantification works as necessary. Also, if you have when p is true, and q is false, and have that (p$rightarrow$p) and (p$rightarrow$(q$rightarrow$p)) are both tautologies, and there exist only two truth values, "true" and "false", then all rows of the truth table follow.
– Doug Spoonwood
Jul 22 at 8:15
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A fundamental distinction here is that an argument is often about many similar situations that the argument treats uniformly. Often this can be seen because some variable appears both in the premise and the conclusion:
Assume that $x$ is an even number.
Then $x^2$ is an even number.
Here we can see the $x$ on both sides, and we can apply your definition directly: The argument is valid because it is impossible that $x$ is even and $x^2$ is not even.
(Sometimes the shared variable is hidden in natural language: "If a number is even, then its square is even too". That makes no formal difference).
In contrast the connective $supset$ and its truth table is in principle always something you apply to a single situations. The formula
$$ xtext is even supset x^2text is even $$
does not have a truth value until you plug in a concrete value for $x$. Once you do that, you can start applying the truth table:
$$ 4text is even supset 16text is even $$
is true due to the first line of the truth table, and
$$ 5text is even supset 25text is even $$
is true due to the last line of the truth table.
So what we can say is that
- The argument "If $p(x)$ then $q(x)$" is valid, and
- The formula "$p(x)supset q(x)$" is always true, meaning for every value of $x$.
are two ways of making the same claim. With quantifiers we can express the second of these more succinctly as
- The formula "$forall x(p(x)supset q(x)) $" is true. (No "always" here).
The truth table for $supset$ can really only be understood when one knows that it is designed to be used together with an (implicit or explicit) quantification over all the situations one is interested in.
Unfortunately a large majority of introductory texts expect students to understand $supset$ before they even begin to speak about quantifiers, and they spend huge amounts of verbiage trying to cajole the reader into thinking it makes sense even then. It doesn't really.
answered Jul 22 at 0:40
Henning Makholm
226k16290518
A fundamental distinction here is that an argument is often about many similar situations that the argument treats uniformly. Often this can be seen because some variable appears both in the premise and the conclusion:
Assume that $x$ is an even number.
Then $x^2$ is an even number.
Here we can see the $x$ on both sides, and we can apply your definition directly: The argument is valid because it is impossible that $x$ is even and $x^2$ is not even.
(Sometimes the shared variable is hidden in natural language: "If a number is even, then its square is even too". That makes no formal difference).
In contrast the connective $supset$ and its truth table is in principle always something you apply to a single situations. The formula
$$ xtext is even supset x^2text is even $$
does not have a truth value until you plug in a concrete value for $x$. Once you do that, you can start applying the truth table:
$$ 4text is even supset 16text is even $$
is true due to the first line of the truth table, and
$$ 5text is even supset 25text is even $$
is true due to the last line of the truth table.
So what we can say is that
- The argument "If $p(x)$ then $q(x)$" is valid, and
- The formula "$p(x)supset q(x)$" is always true, meaning for every value of $x$.
are two ways of making the same claim. With quantifiers we can express the second of these more succinctly as
- The formula "$forall x(p(x)supset q(x)) $" is true. (No "always" here).
The truth table for $supset$ can really only be understood when one knows that it is designed to be used together with an (implicit or explicit) quantification over all the situations one is interested in.
Unfortunately a large majority of introductory texts expect students to understand $supset$ before they even begin to speak about quantifiers, and they spend huge amounts of verbiage trying to cajole the reader into thinking it makes sense even then. It doesn't really.
answered Jul 22 at 0:40
Henning Makholm
226k16290518
answered Jul 22 at 0:40
Henning Makholm
226k16290518
answered Jul 22 at 0:40
Henning Makholm
226k16290518
answered Jul 22 at 0:40
Henning Makholm
226k16290518
226k16290518
If "all men are mortal" and "Henning is a man", then "Henning is mortal" works fine with the conditional. It doesn't require any sort of modern universal or existential quantification. So, I don't know why you've insisted that quantification works as necessary. Also, if you have when p is true, and q is false, and have that (p$rightarrow$p) and (p$rightarrow$(q$rightarrow$p)) are both tautologies, and there exist only two truth values, "true" and "false", then all rows of the truth table follow.
– Doug Spoonwood
Jul 22 at 8:15
add a comment |Â
If "all men are mortal" and "Henning is a man", then "Henning is mortal" works fine with the conditional. It doesn't require any sort of modern universal or existential quantification. So, I don't know why you've insisted that quantification works as necessary. Also, if you have when p is true, and q is false, and have that (p$rightarrow$p) and (p$rightarrow$(q$rightarrow$p)) are both tautologies, and there exist only two truth values, "true" and "false", then all rows of the truth table follow.
– Doug Spoonwood
Jul 22 at 8:15
If "all men are mortal" and "Henning is a man", then "Henning is mortal" works fine with the conditional. It doesn't require any sort of modern universal or existential quantification. So, I don't know why you've insisted that quantification works as necessary. Also, if you have when p is true, and q is false, and have that (p$rightarrow$p) and (p$rightarrow$(q$rightarrow$p)) are both tautologies, and there exist only two truth values, "true" and "false", then all rows of the truth table follow.
– Doug Spoonwood
Jul 22 at 8:15
If "all men are mortal" and "Henning is a man", then "Henning is mortal" works fine with the conditional. It doesn't require any sort of modern universal or existential quantification. So, I don't know why you've insisted that quantification works as necessary. Also, if you have when p is true, and q is false, and have that (p$rightarrow$p) and (p$rightarrow$(q$rightarrow$p)) are both tautologies, and there exist only two truth values, "true" and "false", then all rows of the truth table follow.
– Doug Spoonwood
Jul 22 at 8:15
add a comment |Â
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1
See my formal proof at dcproof.com/DeriveImplies.html There I derive the usual "definition" of material implication using what I believe to be elementary, self-evident rules of inference in natural deduction. Also see my recent blog posting dcproof.com/IfPigsCanFly.html
– Dan Christensen
Jul 22 at 3:51