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What is a natural deduction proof from ~(A↔B) to ~(A→B)?

help with deductive proofUsing natural deduction rules give a formal proofIntroductory Natural Deduction QuestionGiven P ∨ ¬ P prove (P → Q) → ((¬ P → Q) → Q) by natural deductionHow to construct a counter-model of □P --> □◊P in T and K?Natural deduction proof help!Any solution to prove (∀x)(∃y)(Fx & Gy) ⊢ (∃y)(∀x)(Fx & Gy) with natural deduction?How would i go about using natural deduction to prove this argument is valid?Axiomatic proof of ⊢ □P → □◇□P in S4S5 proof of ⊢◻(◻P→◻Q)∨◻(◻Q→◻P)

1

It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.

logic deduction

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edited 8 hours ago

Frank Hubeny

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asked 9 hours ago

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  I made an edit which you may roll back or continue editing. Welcome!
  
  – Frank Hubeny
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It is not intuitively correct; Two predicates not being equivalent does not prohibit one from implying the other.
  
  – Graham Kemp
  2 hours ago
  

  
  
  
  

add a comment | 

1

It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.

logic deduction

share|improve this question

edited 8 hours ago

Frank Hubeny

11.7k51564

asked 9 hours ago

zzzzzzzz

91

New contributor

zzzz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I made an edit which you may roll back or continue editing. Welcome!
  
  – Frank Hubeny
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It is not intuitively correct; Two predicates not being equivalent does not prohibit one from implying the other.
  
  – Graham Kemp
  2 hours ago
  

  
  
  
  

add a comment | 

It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.

logic deduction

share|improve this question

edited 8 hours ago

Frank Hubeny

11.7k51564

asked 9 hours ago

zzzzzzzz

91

New contributor

zzzz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question

edited 8 hours ago

Frank Hubeny

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asked 9 hours ago

zzzzzzzz

91

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share|improve this question

edited 8 hours ago

Frank Hubeny

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asked 9 hours ago

zzzzzzzz

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edited 8 hours ago

Frank Hubeny

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edited 8 hours ago

Frank Hubeny

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asked 9 hours ago

zzzzzzzz

91

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asked 9 hours ago

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3

The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:

If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.

Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.

---

Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html

share|improve this answer

answered 8 hours ago

Frank HubenyFrank Hubeny

11.7k51564

add a comment | 

2

You can't derive ~(A→B) from ~(A↔B).

Consider:

A = I'm in Paris.
B = I'm in France.

~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).

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answered 5 hours ago

EliranEliran

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3

The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:

If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.

Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.

---

Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html

share|improve this answer

answered 8 hours ago

Frank HubenyFrank Hubeny

11.7k51564

add a comment | 

3

The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:

If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.

Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.

---

Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html

share|improve this answer

answered 8 hours ago

Frank HubenyFrank Hubeny

11.7k51564

add a comment | 

The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:

If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.

Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.

---

Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html

share|improve this answer

answered 8 hours ago

Frank HubenyFrank Hubeny

11.7k51564

share|improve this answer

answered 8 hours ago

Frank HubenyFrank Hubeny

11.7k51564

answered 8 hours ago

Frank HubenyFrank Hubeny

11.7k51564

answered 8 hours ago

Frank HubenyFrank Hubeny

11.7k51564

2

You can't derive ~(A→B) from ~(A↔B).

Consider:

A = I'm in Paris.
B = I'm in France.

~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).

share|improve this answer

answered 5 hours ago

EliranEliran

4,52231435

add a comment | 

2

You can't derive ~(A→B) from ~(A↔B).

Consider:

A = I'm in Paris.
B = I'm in France.

~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).

share|improve this answer

answered 5 hours ago

EliranEliran

4,52231435

add a comment | 

You can't derive ~(A→B) from ~(A↔B).

Consider:

A = I'm in Paris.
B = I'm in France.

~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).

share|improve this answer

answered 5 hours ago

EliranEliran

4,52231435

share|improve this answer

answered 5 hours ago

EliranEliran

4,52231435

answered 5 hours ago

EliranEliran

4,52231435

answered 5 hours ago

EliranEliran

4,52231435