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Modus Ponens - Implication vs Disjunction

2

$begingroup$

The Modus Ponens inference rule is generally expressed as:

$$ beginarrayrl
& Prightarrow Q \
& P \
hline
therefore & Qendarray $$

---

Is the below rule also considered to be Modus Ponens?

$$ beginarrayrl
& P lor lnot Q \
& Q \
hline
therefore & Pendarray $$

logic propositional-calculus

share|cite|improve this question

edited Mar 18 at 20:18

Abhilash k

asked Mar 18 at 20:10

Abhilash kAbhilash k

1314

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Translate $P lor lnot Q$ into $lnot Q lor P$ (by commutativity of $lor$) and then to an implication
  $endgroup$
  – Bernard Massé
  Mar 18 at 20:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! I don't know why I didn't think of that! Thanks!
  $endgroup$
  – Abhilash k
  Mar 18 at 20:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You didn't think of it because you are new to the game. By practice and by asking questions when you are stuck, you'll become an expert or at least a good player in the game of logic.
  $endgroup$
  – Bernard Massé
  Mar 18 at 20:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes; it is a particular case of the Resolution rule.
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 19 at 8:10
  

  
  
  
  

add a comment | 

2

$begingroup$

The Modus Ponens inference rule is generally expressed as:

$$ beginarrayrl
& Prightarrow Q \
& P \
hline
therefore & Qendarray $$

---

Is the below rule also considered to be Modus Ponens?

$$ beginarrayrl
& P lor lnot Q \
& Q \
hline
therefore & Pendarray $$

logic propositional-calculus

share|cite|improve this question

edited Mar 18 at 20:18

Abhilash k

asked Mar 18 at 20:10

Abhilash kAbhilash k

1314

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Translate $P lor lnot Q$ into $lnot Q lor P$ (by commutativity of $lor$) and then to an implication
  $endgroup$
  – Bernard Massé
  Mar 18 at 20:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! I don't know why I didn't think of that! Thanks!
  $endgroup$
  – Abhilash k
  Mar 18 at 20:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You didn't think of it because you are new to the game. By practice and by asking questions when you are stuck, you'll become an expert or at least a good player in the game of logic.
  $endgroup$
  – Bernard Massé
  Mar 18 at 20:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes; it is a particular case of the Resolution rule.
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 19 at 8:10
  

  
  
  
  

add a comment | 

$begingroup$

The Modus Ponens inference rule is generally expressed as:

$$ beginarrayrl
& Prightarrow Q \
& P \
hline
therefore & Qendarray $$

---

Is the below rule also considered to be Modus Ponens?

$$ beginarrayrl
& P lor lnot Q \
& Q \
hline
therefore & Pendarray $$

logic propositional-calculus

share|cite|improve this question

edited Mar 18 at 20:18

Abhilash k

asked Mar 18 at 20:10

Abhilash kAbhilash k

1314

$endgroup$

share|cite|improve this question

edited Mar 18 at 20:18

Abhilash k

asked Mar 18 at 20:10

Abhilash kAbhilash k

1314

share|cite|improve this question

edited Mar 18 at 20:18

Abhilash k

asked Mar 18 at 20:10

Abhilash kAbhilash k

1314

asked Mar 18 at 20:10

Abhilash kAbhilash k

1314

asked Mar 18 at 20:10

Abhilash kAbhilash k

1314

2 Answers
2

active

oldest

votes

2

$begingroup$

Yes.

Thanks to @Bernard Massé for pointing me in the right direction.

Here's the proof:

1. $(P lor lnot Q)$ can be written as $ (lnot Q lor P) $ - Commutative Property




2. $ (lnot Q lor P) $ can be written as $ (Q rightarrow P) $ - Material Implication




3. By Modus Ponens :

$$ beginarrayrl
& Qrightarrow P \
& Q \
hline
therefore & Pendarray $$
This is equivalent to

$$ beginarrayrl
& P lor lnot Q \
& Q \
hline
therefore & Pendarray $$

share|cite|improve this answer

answered Mar 18 at 20:36

Abhilash kAbhilash k

1314

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  This is an answer to a slightly different question than what you asked which is whether the latter rule would be considered Modus Ponens. I, personally, would consider it (a special case of) resolution. For me to call it Modus Ponens, I would have had to define $Pto Q$ as a syntactic abbrevation for $neg Plor Q$. Part of the reason I have this view is that the "Material Implication" equivalence is not derivable in constructive logic while both of the rules you list are, so these rules are talking about different things.
  $endgroup$
  – Derek Elkins
  Mar 19 at 0:03
  

  
  
  
  

add a comment | 

1

$begingroup$

I would consider it an application of Disjunctive Syllogism, which is typically stated as:

$P lor Q$

$neg P$

$therefore Q$

Of course, that is not exactly the same pattern, but the basic idea of Disjunctive Syllogism is that you have two options ... but it isn't one of them, and therefore you are left with the other one. Your argument is like that too: it is either P, or $neg Q$, but given $Q$ it is not $neg Q$, and so you are left with $P$

share|cite|improve this answer

answered Mar 21 at 1:50

Bram28Bram28

63.9k44793

$endgroup$

add a comment | 

Your Answer

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2

$begingroup$

Yes.

Thanks to @Bernard Massé for pointing me in the right direction.

Here's the proof:

1. $(P lor lnot Q)$ can be written as $ (lnot Q lor P) $ - Commutative Property




2. $ (lnot Q lor P) $ can be written as $ (Q rightarrow P) $ - Material Implication




3. By Modus Ponens :

$$ beginarrayrl
& Qrightarrow P \
& Q \
hline
therefore & Pendarray $$
This is equivalent to

$$ beginarrayrl
& P lor lnot Q \
& Q \
hline
therefore & Pendarray $$

share|cite|improve this answer

answered Mar 18 at 20:36

Abhilash kAbhilash k

1314

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  This is an answer to a slightly different question than what you asked which is whether the latter rule would be considered Modus Ponens. I, personally, would consider it (a special case of) resolution. For me to call it Modus Ponens, I would have had to define $Pto Q$ as a syntactic abbrevation for $neg Plor Q$. Part of the reason I have this view is that the "Material Implication" equivalence is not derivable in constructive logic while both of the rules you list are, so these rules are talking about different things.
  $endgroup$
  – Derek Elkins
  Mar 19 at 0:03
  

  
  
  
  

add a comment | 

2

$begingroup$

Yes.

Thanks to @Bernard Massé for pointing me in the right direction.

Here's the proof:

1. $(P lor lnot Q)$ can be written as $ (lnot Q lor P) $ - Commutative Property




2. $ (lnot Q lor P) $ can be written as $ (Q rightarrow P) $ - Material Implication




3. By Modus Ponens :

$$ beginarrayrl
& Qrightarrow P \
& Q \
hline
therefore & Pendarray $$
This is equivalent to

$$ beginarrayrl
& P lor lnot Q \
& Q \
hline
therefore & Pendarray $$

share|cite|improve this answer

answered Mar 18 at 20:36

Abhilash kAbhilash k

1314

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  This is an answer to a slightly different question than what you asked which is whether the latter rule would be considered Modus Ponens. I, personally, would consider it (a special case of) resolution. For me to call it Modus Ponens, I would have had to define $Pto Q$ as a syntactic abbrevation for $neg Plor Q$. Part of the reason I have this view is that the "Material Implication" equivalence is not derivable in constructive logic while both of the rules you list are, so these rules are talking about different things.
  $endgroup$
  – Derek Elkins
  Mar 19 at 0:03
  

  
  
  
  

add a comment | 

$begingroup$

Yes.

Thanks to @Bernard Massé for pointing me in the right direction.

Here's the proof:

1. $(P lor lnot Q)$ can be written as $ (lnot Q lor P) $ - Commutative Property




2. $ (lnot Q lor P) $ can be written as $ (Q rightarrow P) $ - Material Implication




3. By Modus Ponens :

$$ beginarrayrl
& Qrightarrow P \
& Q \
hline
therefore & Pendarray $$
This is equivalent to

$$ beginarrayrl
& P lor lnot Q \
& Q \
hline
therefore & Pendarray $$

share|cite|improve this answer

answered Mar 18 at 20:36

Abhilash kAbhilash k

1314

$endgroup$

share|cite|improve this answer

answered Mar 18 at 20:36

Abhilash kAbhilash k

1314

answered Mar 18 at 20:36

Abhilash kAbhilash k

1314

answered Mar 18 at 20:36

Abhilash kAbhilash k

1314

1

$begingroup$

I would consider it an application of Disjunctive Syllogism, which is typically stated as:

$P lor Q$

$neg P$

$therefore Q$

Of course, that is not exactly the same pattern, but the basic idea of Disjunctive Syllogism is that you have two options ... but it isn't one of them, and therefore you are left with the other one. Your argument is like that too: it is either P, or $neg Q$, but given $Q$ it is not $neg Q$, and so you are left with $P$

share|cite|improve this answer

answered Mar 21 at 1:50

Bram28Bram28

63.9k44793

$endgroup$

add a comment | 

1

$begingroup$

I would consider it an application of Disjunctive Syllogism, which is typically stated as:

$P lor Q$

$neg P$

$therefore Q$

Of course, that is not exactly the same pattern, but the basic idea of Disjunctive Syllogism is that you have two options ... but it isn't one of them, and therefore you are left with the other one. Your argument is like that too: it is either P, or $neg Q$, but given $Q$ it is not $neg Q$, and so you are left with $P$

share|cite|improve this answer

answered Mar 21 at 1:50

Bram28Bram28

63.9k44793

$endgroup$

add a comment | 

$begingroup$

I would consider it an application of Disjunctive Syllogism, which is typically stated as:

$P lor Q$

$neg P$

$therefore Q$

Of course, that is not exactly the same pattern, but the basic idea of Disjunctive Syllogism is that you have two options ... but it isn't one of them, and therefore you are left with the other one. Your argument is like that too: it is either P, or $neg Q$, but given $Q$ it is not $neg Q$, and so you are left with $P$

share|cite|improve this answer

answered Mar 21 at 1:50

Bram28Bram28

63.9k44793

$endgroup$

share|cite|improve this answer

answered Mar 21 at 1:50

Bram28Bram28

63.9k44793

answered Mar 21 at 1:50

Bram28Bram28

63.9k44793

answered Mar 21 at 1:50

Bram28Bram28

63.9k44793