How to check if compound proposition is contradiction (is always false)?

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I know how to check compound proposition if it is tautology with TautologyQ. But I have not found any function that can check if the compound proposition is contradiction (is always false). Is there any similar way (like with tautology, without generating truth tables) to check if a compound proposition is a contradiction or not?

For example, how can I check if compound proposition p && !p is a contradiction or not?

boolean-computation logic

share|improve this question

edited Aug 23 at 17:35

asked Aug 22 at 16:28

vasili111

1496

add a comment | 

up vote
5
down vote

favorite

I know how to check compound proposition if it is tautology with TautologyQ. But I have not found any function that can check if the compound proposition is contradiction (is always false). Is there any similar way (like with tautology, without generating truth tables) to check if a compound proposition is a contradiction or not?

For example, how can I check if compound proposition p && !p is a contradiction or not?

boolean-computation logic

share|improve this question

edited Aug 23 at 17:35

asked Aug 22 at 16:28

vasili111

1496

add a comment | 

up vote
5
down vote

favorite

up vote
5
down vote

favorite

I know how to check compound proposition if it is tautology with TautologyQ. But I have not found any function that can check if the compound proposition is contradiction (is always false). Is there any similar way (like with tautology, without generating truth tables) to check if a compound proposition is a contradiction or not?

For example, how can I check if compound proposition p && !p is a contradiction or not?

boolean-computation logic

share|improve this question

edited Aug 23 at 17:35

asked Aug 22 at 16:28

vasili111

1496

I know how to check compound proposition if it is tautology with TautologyQ. But I have not found any function that can check if the compound proposition is contradiction (is always false). Is there any similar way (like with tautology, without generating truth tables) to check if a compound proposition is a contradiction or not?

For example, how can I check if compound proposition p && !p is a contradiction or not?

boolean-computation logic

share|improve this question

edited Aug 23 at 17:35

asked Aug 22 at 16:28

vasili111

1496

share|improve this question

share|improve this question

edited Aug 23 at 17:35

edited Aug 23 at 17:35

edited Aug 23 at 17:35

asked Aug 22 at 16:28

vasili111

1496

asked Aug 22 at 16:28

vasili111

1496

asked Aug 22 at 16:28

vasili111

1496

1496

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
7
down vote



accepted

The converse of tautology (negation of tautology) is a contradiction. More about it here: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology

So to find out if the proposition is a contradiction we can negate the proposition and after check the result if it is the tautology. If the output is True it means that the proposition is contradiction because as we mentioned above the negation of a contradiction is a tautology. If the output is False, that means that the proposition is not contradiction and it can be tautology or contingency.

For example, if we want to check if p && ! p is a contradiction (which it is) we use code:

TautologyQ[Not[p && ! p], p]

Output:

True

share|improve this answer

edited Aug 23 at 13:24

vasili111

1496

answered Aug 22 at 17:00

Henrik Schumacher

36.2k249102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does this checks "if the proposition is a contradiction" or it checks "if the proposition is not a tautology"?
  – vasili111
  Aug 22 at 18:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It checks that the converse of the propositions is a tautology. In my understanding, that means that the proposition is contradiction.
  – Henrik Schumacher
  Aug 22 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you are right: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology
  – vasili111
  Aug 22 at 18:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If f is a boolean-valued function, then f(x1, x2, ..., xn) is true for all possible values of the arguments if and only if ¬f(x1, x2, ..., xn) is false for all possible values of the arguments. Hence, Henrik's approach is valid.
  – Andreas Rejbrand
  Aug 23 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited the answer and added more clarification and information from comments. Is it ok?
  – vasili111
  Aug 23 at 13:13
  

  
  
  
  

 | 
show 2 more comments

up vote
7
down vote

Resolve[Exists[p, p && ! p == True]]

(*

False

*)

You can interpret this result as: "There does not exist any p such that the statement p && !p is True.

Background example:

Resolve[Exists[p, q, (p && ! p == True || q == True)]]

(*

True

*)

You can interpret this result to say:

It is true that there exists some p and q such that both p and !p are True OR q is True. The key claus here is the OR. Certainly there exists some q for which q is True.

share|improve this answer

edited Aug 22 at 17:02

answered Aug 22 at 16:36

David G. Stork

21.3k11646

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your example, I can substitute p, p && ! p for any other compound proposition to check for a contradiction, right? For example, for a||bThere sould be Resolve[Exists[a, b, a && b == True]] ?
  – vasili111
  Aug 22 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also if False it means contradiction and if True it means that it is not a contradiction?
  – vasili111
  Aug 22 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  False in my solution example means there does not exist any p such that p and not p is True.
  – David G. Stork
  Aug 22 at 17:01
  

  
  
  
  

add a comment | 

up vote
1
down vote

I also found another way:

"In a complete logic, a formula is contradictory if and only if it is unsatisfiable". Source

In that case, we can check if the proposition is satisfiable and after negate the result. If the output is Truethat means that the original statement was unsatisfiable which also means that it was contradiction. If the output was False it means that the result was satisfiable, which means that it was tautology or contingency.

---

For example, lets check if p && !p is contradiction:

Not[SatisfiableQ[p && ! p, p]]

output:

True

It means that p && !p is contradiction.

---

Lets check if p || ! p is contradiction:

Not[SatisfiableQ[p || ! p, p]]

output:

False

It means that p || ! p is not contradiction. The output does not say that but it is tautology.

---

Lets check if p && q is contradiction:

Not[SatisfiableQ[p && q, p, q]]

output:

False

It means that p && q is not contradiction. The output does not say that but it is contingency.

share|improve this answer

answered Aug 23 at 13:52

vasili111

1496

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That makes sense because of: $neg (exists p colon text$A(p)$ is true)$ is equivalent to $(forall p colon neg (text$A(p)$ is true))$. SatisfiableQ is essentially the $exists$-quantor.
  – Henrik Schumacher
  Aug 23 at 13:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher Thank you.
  – vasili111
  Aug 23 at 13:58
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

The converse of tautology (negation of tautology) is a contradiction. More about it here: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology

So to find out if the proposition is a contradiction we can negate the proposition and after check the result if it is the tautology. If the output is True it means that the proposition is contradiction because as we mentioned above the negation of a contradiction is a tautology. If the output is False, that means that the proposition is not contradiction and it can be tautology or contingency.

For example, if we want to check if p && ! p is a contradiction (which it is) we use code:

TautologyQ[Not[p && ! p], p]

Output:

True

share|improve this answer

edited Aug 23 at 13:24

vasili111

1496

answered Aug 22 at 17:00

Henrik Schumacher

36.2k249102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does this checks "if the proposition is a contradiction" or it checks "if the proposition is not a tautology"?
  – vasili111
  Aug 22 at 18:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It checks that the converse of the propositions is a tautology. In my understanding, that means that the proposition is contradiction.
  – Henrik Schumacher
  Aug 22 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you are right: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology
  – vasili111
  Aug 22 at 18:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If f is a boolean-valued function, then f(x1, x2, ..., xn) is true for all possible values of the arguments if and only if ¬f(x1, x2, ..., xn) is false for all possible values of the arguments. Hence, Henrik's approach is valid.
  – Andreas Rejbrand
  Aug 23 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited the answer and added more clarification and information from comments. Is it ok?
  – vasili111
  Aug 23 at 13:13
  

  
  
  
  

 | 
show 2 more comments

up vote
7
down vote



accepted

The converse of tautology (negation of tautology) is a contradiction. More about it here: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology

So to find out if the proposition is a contradiction we can negate the proposition and after check the result if it is the tautology. If the output is True it means that the proposition is contradiction because as we mentioned above the negation of a contradiction is a tautology. If the output is False, that means that the proposition is not contradiction and it can be tautology or contingency.

For example, if we want to check if p && ! p is a contradiction (which it is) we use code:

TautologyQ[Not[p && ! p], p]

Output:

True

share|improve this answer

edited Aug 23 at 13:24

vasili111

1496

answered Aug 22 at 17:00

Henrik Schumacher

36.2k249102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does this checks "if the proposition is a contradiction" or it checks "if the proposition is not a tautology"?
  – vasili111
  Aug 22 at 18:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It checks that the converse of the propositions is a tautology. In my understanding, that means that the proposition is contradiction.
  – Henrik Schumacher
  Aug 22 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you are right: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology
  – vasili111
  Aug 22 at 18:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If f is a boolean-valued function, then f(x1, x2, ..., xn) is true for all possible values of the arguments if and only if ¬f(x1, x2, ..., xn) is false for all possible values of the arguments. Hence, Henrik's approach is valid.
  – Andreas Rejbrand
  Aug 23 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited the answer and added more clarification and information from comments. Is it ok?
  – vasili111
  Aug 23 at 13:13
  

  
  
  
  

 | 
show 2 more comments

up vote
7
down vote



accepted

up vote
7
down vote



accepted

The converse of tautology (negation of tautology) is a contradiction. More about it here: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology

So to find out if the proposition is a contradiction we can negate the proposition and after check the result if it is the tautology. If the output is True it means that the proposition is contradiction because as we mentioned above the negation of a contradiction is a tautology. If the output is False, that means that the proposition is not contradiction and it can be tautology or contingency.

For example, if we want to check if p && ! p is a contradiction (which it is) we use code:

TautologyQ[Not[p && ! p], p]

Output:

True

share|improve this answer

edited Aug 23 at 13:24

vasili111

1496

answered Aug 22 at 17:00

Henrik Schumacher

36.2k249102

The converse of tautology (negation of tautology) is a contradiction. More about it here: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology

So to find out if the proposition is a contradiction we can negate the proposition and after check the result if it is the tautology. If the output is True it means that the proposition is contradiction because as we mentioned above the negation of a contradiction is a tautology. If the output is False, that means that the proposition is not contradiction and it can be tautology or contingency.

For example, if we want to check if p && ! p is a contradiction (which it is) we use code:

TautologyQ[Not[p && ! p], p]

Output:

True

share|improve this answer

edited Aug 23 at 13:24

vasili111

1496

answered Aug 22 at 17:00

Henrik Schumacher

36.2k249102

share|improve this answer

share|improve this answer

edited Aug 23 at 13:24

vasili111

1496

edited Aug 23 at 13:24

vasili111

1496

edited Aug 23 at 13:24

vasili111

1496

1496

answered Aug 22 at 17:00

Henrik Schumacher

36.2k249102

answered Aug 22 at 17:00

Henrik Schumacher

36.2k249102

answered Aug 22 at 17:00

Henrik Schumacher

36.2k249102

36.2k249102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does this checks "if the proposition is a contradiction" or it checks "if the proposition is not a tautology"?
  – vasili111
  Aug 22 at 18:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It checks that the converse of the propositions is a tautology. In my understanding, that means that the proposition is contradiction.
  – Henrik Schumacher
  Aug 22 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you are right: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology
  – vasili111
  Aug 22 at 18:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If f is a boolean-valued function, then f(x1, x2, ..., xn) is true for all possible values of the arguments if and only if ¬f(x1, x2, ..., xn) is false for all possible values of the arguments. Hence, Henrik's approach is valid.
  – Andreas Rejbrand
  Aug 23 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited the answer and added more clarification and information from comments. Is it ok?
  – vasili111
  Aug 23 at 13:13
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does this checks "if the proposition is a contradiction" or it checks "if the proposition is not a tautology"?
  – vasili111
  Aug 22 at 18:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It checks that the converse of the propositions is a tautology. In my understanding, that means that the proposition is contradiction.
  – Henrik Schumacher
  Aug 22 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you are right: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology
  – vasili111
  Aug 22 at 18:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If f is a boolean-valued function, then f(x1, x2, ..., xn) is true for all possible values of the arguments if and only if ¬f(x1, x2, ..., xn) is false for all possible values of the arguments. Hence, Henrik's approach is valid.
  – Andreas Rejbrand
  Aug 23 at 6:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited the answer and added more clarification and information from comments. Is it ok?
  – vasili111
  Aug 23 at 13:13
  

  
  
  
  

Does this checks "if the proposition is a contradiction" or it checks "if the proposition is not a tautology"?
– vasili111
Aug 22 at 18:05

Does this checks "if the proposition is a contradiction" or it checks "if the proposition is not a tautology"?
– vasili111
Aug 22 at 18:05

1

1

It checks that the converse of the propositions is a tautology. In my understanding, that means that the proposition is contradiction.
– Henrik Schumacher
Aug 22 at 18:07

It checks that the converse of the propositions is a tautology. In my understanding, that means that the proposition is contradiction.
– Henrik Schumacher
Aug 22 at 18:07

I think you are right: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology
– vasili111
Aug 22 at 18:24

I think you are right: proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology
– vasili111
Aug 22 at 18:24

1

1

If f is a boolean-valued function, then f(x1, x2, ..., xn) is true for all possible values of the arguments if and only if ¬f(x1, x2, ..., xn) is false for all possible values of the arguments. Hence, Henrik's approach is valid.
– Andreas Rejbrand
Aug 23 at 6:16

If f is a boolean-valued function, then f(x1, x2, ..., xn) is true for all possible values of the arguments if and only if ¬f(x1, x2, ..., xn) is false for all possible values of the arguments. Hence, Henrik's approach is valid.
– Andreas Rejbrand
Aug 23 at 6:16

I edited the answer and added more clarification and information from comments. Is it ok?
– vasili111
Aug 23 at 13:13

I edited the answer and added more clarification and information from comments. Is it ok?
– vasili111
Aug 23 at 13:13

 | 
show 2 more comments

up vote
7
down vote

Resolve[Exists[p, p && ! p == True]]

(*

False

*)

You can interpret this result as: "There does not exist any p such that the statement p && !p is True.

Background example:

Resolve[Exists[p, q, (p && ! p == True || q == True)]]

(*

True

*)

You can interpret this result to say:

It is true that there exists some p and q such that both p and !p are True OR q is True. The key claus here is the OR. Certainly there exists some q for which q is True.

share|improve this answer

edited Aug 22 at 17:02

answered Aug 22 at 16:36

David G. Stork

21.3k11646

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your example, I can substitute p, p && ! p for any other compound proposition to check for a contradiction, right? For example, for a||bThere sould be Resolve[Exists[a, b, a && b == True]] ?
  – vasili111
  Aug 22 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also if False it means contradiction and if True it means that it is not a contradiction?
  – vasili111
  Aug 22 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  False in my solution example means there does not exist any p such that p and not p is True.
  – David G. Stork
  Aug 22 at 17:01
  

  
  
  
  

add a comment | 

up vote
7
down vote

Resolve[Exists[p, p && ! p == True]]

(*

False

*)

You can interpret this result as: "There does not exist any p such that the statement p && !p is True.

Background example:

Resolve[Exists[p, q, (p && ! p == True || q == True)]]

(*

True

*)

You can interpret this result to say:

It is true that there exists some p and q such that both p and !p are True OR q is True. The key claus here is the OR. Certainly there exists some q for which q is True.

share|improve this answer

edited Aug 22 at 17:02

answered Aug 22 at 16:36

David G. Stork

21.3k11646

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your example, I can substitute p, p && ! p for any other compound proposition to check for a contradiction, right? For example, for a||bThere sould be Resolve[Exists[a, b, a && b == True]] ?
  – vasili111
  Aug 22 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also if False it means contradiction and if True it means that it is not a contradiction?
  – vasili111
  Aug 22 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  False in my solution example means there does not exist any p such that p and not p is True.
  – David G. Stork
  Aug 22 at 17:01
  

  
  
  
  

add a comment | 

up vote
7
down vote

up vote
7
down vote

Resolve[Exists[p, p && ! p == True]]

(*

False

*)

You can interpret this result as: "There does not exist any p such that the statement p && !p is True.

Background example:

Resolve[Exists[p, q, (p && ! p == True || q == True)]]

(*

True

*)

You can interpret this result to say:

It is true that there exists some p and q such that both p and !p are True OR q is True. The key claus here is the OR. Certainly there exists some q for which q is True.

share|improve this answer

edited Aug 22 at 17:02

answered Aug 22 at 16:36

David G. Stork

21.3k11646

Resolve[Exists[p, p && ! p == True]]

(*

False

*)

You can interpret this result as: "There does not exist any p such that the statement p && !p is True.

Background example:

Resolve[Exists[p, q, (p && ! p == True || q == True)]]

(*

True

*)

You can interpret this result to say:

It is true that there exists some p and q such that both p and !p are True OR q is True. The key claus here is the OR. Certainly there exists some q for which q is True.

share|improve this answer

edited Aug 22 at 17:02

answered Aug 22 at 16:36

David G. Stork

21.3k11646

share|improve this answer

share|improve this answer

edited Aug 22 at 17:02

edited Aug 22 at 17:02

edited Aug 22 at 17:02

answered Aug 22 at 16:36

David G. Stork

21.3k11646

answered Aug 22 at 16:36

David G. Stork

21.3k11646

answered Aug 22 at 16:36

David G. Stork

21.3k11646

21.3k11646

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your example, I can substitute p, p && ! p for any other compound proposition to check for a contradiction, right? For example, for a||bThere sould be Resolve[Exists[a, b, a && b == True]] ?
  – vasili111
  Aug 22 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also if False it means contradiction and if True it means that it is not a contradiction?
  – vasili111
  Aug 22 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  False in my solution example means there does not exist any p such that p and not p is True.
  – David G. Stork
  Aug 22 at 17:01
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In your example, I can substitute p, p && ! p for any other compound proposition to check for a contradiction, right? For example, for a||bThere sould be Resolve[Exists[a, b, a && b == True]] ?
  – vasili111
  Aug 22 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also if False it means contradiction and if True it means that it is not a contradiction?
  – vasili111
  Aug 22 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  False in my solution example means there does not exist any p such that p and not p is True.
  – David G. Stork
  Aug 22 at 17:01
  

  
  
  
  

In your example, I can substitute p, p && ! p for any other compound proposition to check for a contradiction, right? For example, for a||bThere sould be Resolve[Exists[a, b, a && b == True]] ?
– vasili111
Aug 22 at 16:54

In your example, I can substitute p, p && ! p for any other compound proposition to check for a contradiction, right? For example, for a||bThere sould be Resolve[Exists[a, b, a && b == True]] ?
– vasili111
Aug 22 at 16:54

Also if False it means contradiction and if True it means that it is not a contradiction?
– vasili111
Aug 22 at 16:55

Also if False it means contradiction and if True it means that it is not a contradiction?
– vasili111
Aug 22 at 16:55

False in my solution example means there does not exist any p such that p and not p is True.
– David G. Stork
Aug 22 at 17:01

False in my solution example means there does not exist any p such that p and not p is True.
– David G. Stork
Aug 22 at 17:01

add a comment | 

up vote
1
down vote

I also found another way:

"In a complete logic, a formula is contradictory if and only if it is unsatisfiable". Source

In that case, we can check if the proposition is satisfiable and after negate the result. If the output is Truethat means that the original statement was unsatisfiable which also means that it was contradiction. If the output was False it means that the result was satisfiable, which means that it was tautology or contingency.

---

For example, lets check if p && !p is contradiction:

Not[SatisfiableQ[p && ! p, p]]

output:

True

It means that p && !p is contradiction.

---

Lets check if p || ! p is contradiction:

Not[SatisfiableQ[p || ! p, p]]

output:

False

It means that p || ! p is not contradiction. The output does not say that but it is tautology.

---

Lets check if p && q is contradiction:

Not[SatisfiableQ[p && q, p, q]]

output:

False

It means that p && q is not contradiction. The output does not say that but it is contingency.

share|improve this answer

answered Aug 23 at 13:52

vasili111

1496

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That makes sense because of: $neg (exists p colon text$A(p)$ is true)$ is equivalent to $(forall p colon neg (text$A(p)$ is true))$. SatisfiableQ is essentially the $exists$-quantor.
  – Henrik Schumacher
  Aug 23 at 13:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher Thank you.
  – vasili111
  Aug 23 at 13:58
  

  
  
  
  

add a comment | 

up vote
1
down vote

I also found another way:

"In a complete logic, a formula is contradictory if and only if it is unsatisfiable". Source

In that case, we can check if the proposition is satisfiable and after negate the result. If the output is Truethat means that the original statement was unsatisfiable which also means that it was contradiction. If the output was False it means that the result was satisfiable, which means that it was tautology or contingency.

---

For example, lets check if p && !p is contradiction:

Not[SatisfiableQ[p && ! p, p]]

output:

True

It means that p && !p is contradiction.

---

Lets check if p || ! p is contradiction:

Not[SatisfiableQ[p || ! p, p]]

output:

False

It means that p || ! p is not contradiction. The output does not say that but it is tautology.

---

Lets check if p && q is contradiction:

Not[SatisfiableQ[p && q, p, q]]

output:

False

It means that p && q is not contradiction. The output does not say that but it is contingency.

share|improve this answer

answered Aug 23 at 13:52

vasili111

1496

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That makes sense because of: $neg (exists p colon text$A(p)$ is true)$ is equivalent to $(forall p colon neg (text$A(p)$ is true))$. SatisfiableQ is essentially the $exists$-quantor.
  – Henrik Schumacher
  Aug 23 at 13:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher Thank you.
  – vasili111
  Aug 23 at 13:58
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

I also found another way:

"In a complete logic, a formula is contradictory if and only if it is unsatisfiable". Source

In that case, we can check if the proposition is satisfiable and after negate the result. If the output is Truethat means that the original statement was unsatisfiable which also means that it was contradiction. If the output was False it means that the result was satisfiable, which means that it was tautology or contingency.

---

For example, lets check if p && !p is contradiction:

Not[SatisfiableQ[p && ! p, p]]

output:

True

It means that p && !p is contradiction.

---

Lets check if p || ! p is contradiction:

Not[SatisfiableQ[p || ! p, p]]

output:

False

It means that p || ! p is not contradiction. The output does not say that but it is tautology.

---

Lets check if p && q is contradiction:

Not[SatisfiableQ[p && q, p, q]]

output:

False

It means that p && q is not contradiction. The output does not say that but it is contingency.

share|improve this answer

answered Aug 23 at 13:52

vasili111

1496

I also found another way:

"In a complete logic, a formula is contradictory if and only if it is unsatisfiable". Source

In that case, we can check if the proposition is satisfiable and after negate the result. If the output is Truethat means that the original statement was unsatisfiable which also means that it was contradiction. If the output was False it means that the result was satisfiable, which means that it was tautology or contingency.

---

For example, lets check if p && !p is contradiction:

Not[SatisfiableQ[p && ! p, p]]

output:

True

It means that p && !p is contradiction.

---

Lets check if p || ! p is contradiction:

Not[SatisfiableQ[p || ! p, p]]

output:

False

It means that p || ! p is not contradiction. The output does not say that but it is tautology.

---

Lets check if p && q is contradiction:

Not[SatisfiableQ[p && q, p, q]]

output:

False

It means that p && q is not contradiction. The output does not say that but it is contingency.

share|improve this answer

answered Aug 23 at 13:52

vasili111

1496

share|improve this answer

share|improve this answer

answered Aug 23 at 13:52

vasili111

1496

answered Aug 23 at 13:52

vasili111

1496

answered Aug 23 at 13:52

vasili111

1496

1496

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That makes sense because of: $neg (exists p colon text$A(p)$ is true)$ is equivalent to $(forall p colon neg (text$A(p)$ is true))$. SatisfiableQ is essentially the $exists$-quantor.
  – Henrik Schumacher
  Aug 23 at 13:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher Thank you.
  – vasili111
  Aug 23 at 13:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That makes sense because of: $neg (exists p colon text$A(p)$ is true)$ is equivalent to $(forall p colon neg (text$A(p)$ is true))$. SatisfiableQ is essentially the $exists$-quantor.
  – Henrik Schumacher
  Aug 23 at 13:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenrikSchumacher Thank you.
  – vasili111
  Aug 23 at 13:58
  

  
  
  
  

That makes sense because of: $neg (exists p colon text$A(p)$ is true)$ is equivalent to $(forall p colon neg (text$A(p)$ is true))$. SatisfiableQ is essentially the $exists$-quantor.
– Henrik Schumacher
Aug 23 at 13:57

That makes sense because of: $neg (exists p colon text$A(p)$ is true)$ is equivalent to $(forall p colon neg (text$A(p)$ is true))$. SatisfiableQ is essentially the $exists$-quantor.
– Henrik Schumacher
Aug 23 at 13:57

@HenrikSchumacher Thank you.
– vasili111
Aug 23 at 13:58

@HenrikSchumacher Thank you.
– vasili111
Aug 23 at 13:58

add a comment | 

 

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