Mixing
What rule allows the separation of $ p rightarrow (a lor c)$?
$begingroup$
I have seen this in some examples of proofs where implication $ p rightarrow (a lor c)$ is separated, example:
- $ p rightarrow (a lor c)$
- $ p $
- $ (a lor c) $
What is the rule (law of inference?) that allows for this separation?
proof-writing propositional-calculus
asked Jan 28 at 13:44
SamSam
44418
$endgroup$
add a comment |
$begingroup$
I have seen this in some examples of proofs where implication $ p rightarrow (a lor c)$ is separated, example:
- $ p rightarrow (a lor c)$
- $ p $
- $ (a lor c) $
What is the rule (law of inference?) that allows for this separation?
proof-writing propositional-calculus
asked Jan 28 at 13:44
SamSam
44418
$endgroup$
4
$begingroup$
Modus ponens I think. If $pto q$ and $p$ are true then $q$ is true.
$endgroup$
– drhab
Jan 28 at 13:47
add a comment |
$begingroup$
I have seen this in some examples of proofs where implication $ p rightarrow (a lor c)$ is separated, example:
- $ p rightarrow (a lor c)$
- $ p $
- $ (a lor c) $
What is the rule (law of inference?) that allows for this separation?
proof-writing propositional-calculus
asked Jan 28 at 13:44
SamSam
44418
$endgroup$
I have seen this in some examples of proofs where implication $ p rightarrow (a lor c)$ is separated, example:
- $ p rightarrow (a lor c)$
- $ p $
- $ (a lor c) $
What is the rule (law of inference?) that allows for this separation?
proof-writing propositional-calculus
proof-writing propositional-calculus
asked Jan 28 at 13:44
SamSam
44418
asked Jan 28 at 13:44
SamSam
44418
asked Jan 28 at 13:44
SamSam
44418
asked Jan 28 at 13:44
SamSam
44418
asked Jan 28 at 13:44
SamSam
44418
44418
4
$begingroup$
Modus ponens I think. If $pto q$ and $p$ are true then $q$ is true.
$endgroup$
– drhab
Jan 28 at 13:47
add a comment |
4
$begingroup$
Modus ponens I think. If $pto q$ and $p$ are true then $q$ is true.
$endgroup$
– drhab
Jan 28 at 13:47
4
4
$begingroup$
Modus ponens I think. If $pto q$ and $p$ are true then $q$ is true.
$endgroup$
– drhab
Jan 28 at 13:47
$begingroup$
Modus ponens I think. If $pto q$ and $p$ are true then $q$ is true.
$endgroup$
– drhab
Jan 28 at 13:47
add a comment |
1 Answer
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$begingroup$
As @drhab mentioned, the answer is Modus ponens, in general, which has the form
begin{align*}
pto &q\
&p \
therefore ; &q.
end{align*}
In your case, $p=p,$ and $q=alor c$. Here, you need to be able to recognize that Modus ponens is really a schema: you can encapsulate any sort of complicated expression in the $p$ or $q$. If the overall structure has this form, then you can deduce $q,$ whatever that is.
It is also good to note that the name of this deductive rule varies depending on which deductive system you're in. It's certainly most widely known as Modus ponens (Latin for "way of putting"), but in the Natural Deduction system, for example, it's known as "$to$ Elimination".
Some logicians think of "Modus ponens" as essentially the rule of deduction, and have gone to great lengths to show that it is sufficient for all deductive purposes. That it may be, but it is also true that if you allow a few more rules, then your proofs can be shorter. That's one of the balancing acts when you're designing a deductive system: fewer axioms seems "prettier" by mathematical standards; on the other hand, with deduction systems being as practical as they are, it's often considered better practice to allow more rules.
answered Jan 28 at 15:06
Adrian KeisterAdrian Keister
5,28171933
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add a comment |
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1 Answer
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$begingroup$
As @drhab mentioned, the answer is Modus ponens, in general, which has the form
begin{align*}
pto &q\
&p \
therefore ; &q.
end{align*}
In your case, $p=p,$ and $q=alor c$. Here, you need to be able to recognize that Modus ponens is really a schema: you can encapsulate any sort of complicated expression in the $p$ or $q$. If the overall structure has this form, then you can deduce $q,$ whatever that is.
It is also good to note that the name of this deductive rule varies depending on which deductive system you're in. It's certainly most widely known as Modus ponens (Latin for "way of putting"), but in the Natural Deduction system, for example, it's known as "$to$ Elimination".
Some logicians think of "Modus ponens" as essentially the rule of deduction, and have gone to great lengths to show that it is sufficient for all deductive purposes. That it may be, but it is also true that if you allow a few more rules, then your proofs can be shorter. That's one of the balancing acts when you're designing a deductive system: fewer axioms seems "prettier" by mathematical standards; on the other hand, with deduction systems being as practical as they are, it's often considered better practice to allow more rules.
answered Jan 28 at 15:06
Adrian KeisterAdrian Keister
5,28171933
$endgroup$
add a comment |
$begingroup$
As @drhab mentioned, the answer is Modus ponens, in general, which has the form
begin{align*}
pto &q\
&p \
therefore ; &q.
end{align*}
In your case, $p=p,$ and $q=alor c$. Here, you need to be able to recognize that Modus ponens is really a schema: you can encapsulate any sort of complicated expression in the $p$ or $q$. If the overall structure has this form, then you can deduce $q,$ whatever that is.
It is also good to note that the name of this deductive rule varies depending on which deductive system you're in. It's certainly most widely known as Modus ponens (Latin for "way of putting"), but in the Natural Deduction system, for example, it's known as "$to$ Elimination".
Some logicians think of "Modus ponens" as essentially the rule of deduction, and have gone to great lengths to show that it is sufficient for all deductive purposes. That it may be, but it is also true that if you allow a few more rules, then your proofs can be shorter. That's one of the balancing acts when you're designing a deductive system: fewer axioms seems "prettier" by mathematical standards; on the other hand, with deduction systems being as practical as they are, it's often considered better practice to allow more rules.
answered Jan 28 at 15:06
Adrian KeisterAdrian Keister
5,28171933
$endgroup$
add a comment |
$begingroup$
As @drhab mentioned, the answer is Modus ponens, in general, which has the form
begin{align*}
pto &q\
&p \
therefore ; &q.
end{align*}
In your case, $p=p,$ and $q=alor c$. Here, you need to be able to recognize that Modus ponens is really a schema: you can encapsulate any sort of complicated expression in the $p$ or $q$. If the overall structure has this form, then you can deduce $q,$ whatever that is.
It is also good to note that the name of this deductive rule varies depending on which deductive system you're in. It's certainly most widely known as Modus ponens (Latin for "way of putting"), but in the Natural Deduction system, for example, it's known as "$to$ Elimination".
Some logicians think of "Modus ponens" as essentially the rule of deduction, and have gone to great lengths to show that it is sufficient for all deductive purposes. That it may be, but it is also true that if you allow a few more rules, then your proofs can be shorter. That's one of the balancing acts when you're designing a deductive system: fewer axioms seems "prettier" by mathematical standards; on the other hand, with deduction systems being as practical as they are, it's often considered better practice to allow more rules.
answered Jan 28 at 15:06
Adrian KeisterAdrian Keister
5,28171933
$endgroup$
As @drhab mentioned, the answer is Modus ponens, in general, which has the form
begin{align*}
pto &q\
&p \
therefore ; &q.
end{align*}
In your case, $p=p,$ and $q=alor c$. Here, you need to be able to recognize that Modus ponens is really a schema: you can encapsulate any sort of complicated expression in the $p$ or $q$. If the overall structure has this form, then you can deduce $q,$ whatever that is.
It is also good to note that the name of this deductive rule varies depending on which deductive system you're in. It's certainly most widely known as Modus ponens (Latin for "way of putting"), but in the Natural Deduction system, for example, it's known as "$to$ Elimination".
Some logicians think of "Modus ponens" as essentially the rule of deduction, and have gone to great lengths to show that it is sufficient for all deductive purposes. That it may be, but it is also true that if you allow a few more rules, then your proofs can be shorter. That's one of the balancing acts when you're designing a deductive system: fewer axioms seems "prettier" by mathematical standards; on the other hand, with deduction systems being as practical as they are, it's often considered better practice to allow more rules.
answered Jan 28 at 15:06
Adrian KeisterAdrian Keister
5,28171933
answered Jan 28 at 15:06
Adrian KeisterAdrian Keister
5,28171933
answered Jan 28 at 15:06
Adrian KeisterAdrian Keister
5,28171933
answered Jan 28 at 15:06
Adrian KeisterAdrian Keister
5,28171933
5,28171933
add a comment |
add a comment |
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$begingroup$
Modus ponens I think. If $pto q$ and $p$ are true then $q$ is true.
$endgroup$
– drhab
Jan 28 at 13:47