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Fitch Proof Question

Conditional disjunction equivalence proof using FItchFitch Proof - Logic LPL 13.11Fitch Proof - LPL Exercise 8.17fitch proof chapter 13 exercise 13.49Use the Fitch system to prove the tautology (p ∨ ¬p)LPL ( language proof and logic) - FITCH - 14.12Seminal, accessible work on defeasible reasoningIs it possible to define argument validity as a formula?Fitch Biconditional Proof Help?Fitch Proof by Contradiction help

1

I'm having trouble with a proof and I'm not sure if it's valid or not. If it appears to be invalid, we are supposed to assign names to the letters in the proof and check it in a World, but when I do it still checks out.

When I try to prove it in Fitch, I get all the way to the bottom, with (Q^R) as the last step of the subproof and I'm trying to apply -->Intro to join
(NOT)P --> (Q ^ R) but it's not checking out because it says that it needs a support step to be cited.

P --> Q                Premise

R ^ S                  Premise

------------------

(NOT)P --> (Q ^ R)     Goal

logic proof language fitch

share|improve this question

edited 1 hour ago

Frank Hubeny

11k51559

asked 1 hour ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Nope, that is not valid. Are you missing a 'not' before that first P?
  
  – Graham Kemp
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is it not valid because there's no (NOT)P in the premises? Since it's not known? No, it's not missing. The first premise is P --> Q.
  
  – fitch-is-killing-me
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Then you cannot conclude ¬P → (Q ˄ R) from those premises, although you may conclude P → (Q ˄ R)
  
  – Graham Kemp
  1 hour ago
  

  
  
  
  

add a comment | 

1

I'm having trouble with a proof and I'm not sure if it's valid or not. If it appears to be invalid, we are supposed to assign names to the letters in the proof and check it in a World, but when I do it still checks out.

When I try to prove it in Fitch, I get all the way to the bottom, with (Q^R) as the last step of the subproof and I'm trying to apply -->Intro to join
(NOT)P --> (Q ^ R) but it's not checking out because it says that it needs a support step to be cited.

P --> Q                Premise

R ^ S                  Premise

------------------

(NOT)P --> (Q ^ R)     Goal

logic proof language fitch

share|improve this question

edited 1 hour ago

Frank Hubeny

11k51559

asked 1 hour ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Nope, that is not valid. Are you missing a 'not' before that first P?
  
  – Graham Kemp
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is it not valid because there's no (NOT)P in the premises? Since it's not known? No, it's not missing. The first premise is P --> Q.
  
  – fitch-is-killing-me
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Then you cannot conclude ¬P → (Q ˄ R) from those premises, although you may conclude P → (Q ˄ R)
  
  – Graham Kemp
  1 hour ago
  

  
  
  
  

add a comment | 

I'm having trouble with a proof and I'm not sure if it's valid or not. If it appears to be invalid, we are supposed to assign names to the letters in the proof and check it in a World, but when I do it still checks out.

When I try to prove it in Fitch, I get all the way to the bottom, with (Q^R) as the last step of the subproof and I'm trying to apply -->Intro to join
(NOT)P --> (Q ^ R) but it's not checking out because it says that it needs a support step to be cited.

P --> Q                Premise

R ^ S                  Premise

------------------

(NOT)P --> (Q ^ R)     Goal

logic proof language fitch

share|improve this question

edited 1 hour ago

Frank Hubeny

11k51559

asked 1 hour ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

P --> Q                Premise

R ^ S                  Premise

------------------

(NOT)P --> (Q ^ R)     Goal

share|improve this question

edited 1 hour ago

Frank Hubeny

11k51559

asked 1 hour ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

share|improve this question

edited 1 hour ago

Frank Hubeny

11k51559

asked 1 hour ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

edited 1 hour ago

Frank Hubeny

11k51559

edited 1 hour ago

Frank Hubeny

11k51559

asked 1 hour ago

fitch-is-killing-mefitch-is-killing-me

61

New contributor

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

asked 1 hour ago

fitch-is-killing-mefitch-is-killing-me

61

2 Answers
2

active

oldest

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2

P → Q, R ˄ S |- ¬P → (Q ˄ R) is not syntactically valid.

An assumption of ¬P will not allow you to eliminate the conditional in P → Q to derive Q .

Is there a typo?

P → Q, R ˄ S |- P → (Q ˄ R) is valid.

 1|  P → Q

 2|_ R ˄ S

 3|  R               ˄E, 2

 4|  |_ P

 5|  |  Q            →E, 1,4

 6|  |  Q ˄ R        ˄I, 3,5

 7|  P → (Q ˄ R)    →I, 4-6

---

For an argument to be semantically valid, the conclusion must be demonstrably true in all interpretations where the premises are -- it is not enough to find just one. A proof is semantically invalid when the exists some interpretation where all the premises are true, but the conclusion is false. One such counter example is enough to disprove an argument.

In the interpretation of {P=f,Q=f, R=t, S=t}, the premises P → Q, R ˄ S, are both true, but ¬P → (Q ˄ R) is false.

share|improve this answer

edited 53 mins ago

answered 1 hour ago

Graham KempGraham Kemp

1,10919

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No, there's no typo. I imagined that this proof wasn't valid simply because I coulnd't conclude (NOT)P from anywhere. So, when I replace the letters with cube(a) , cube(b), why is it showing, in the world, that the proof is valid? Should I choose names that I know will prove it's invalid?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  

add a comment | 

1

By using a truth table generator one can show that one of the set of valuations for the sentence letters leads to the result being false.

To see this, use this input ((P=>Q)&&(R&&S))=>(~P=>(Q&&R)) in the Stanford Truth Table Tool. When P and Q are false and R and S are true then the conditional with the conjunction of the premises as antecedent and the goal as consequent is false.

This would mean that one could not derive the result.

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That makes sense. So the reason that the proof is invalid, is because out of all the results, there's only 1 that is false? In a valid proof, they should all be true?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That's right. If there were only T values then the statement would be a tautology based on the truth table and a derivation or proof could be found that would be valid.
  
  – Frank Hubeny
  1 hour ago
  

  
  
  
  

add a comment | 

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2

P → Q, R ˄ S |- ¬P → (Q ˄ R) is not syntactically valid.

An assumption of ¬P will not allow you to eliminate the conditional in P → Q to derive Q .

Is there a typo?

P → Q, R ˄ S |- P → (Q ˄ R) is valid.

 1|  P → Q

 2|_ R ˄ S

 3|  R               ˄E, 2

 4|  |_ P

 5|  |  Q            →E, 1,4

 6|  |  Q ˄ R        ˄I, 3,5

 7|  P → (Q ˄ R)    →I, 4-6

---

For an argument to be semantically valid, the conclusion must be demonstrably true in all interpretations where the premises are -- it is not enough to find just one. A proof is semantically invalid when the exists some interpretation where all the premises are true, but the conclusion is false. One such counter example is enough to disprove an argument.

In the interpretation of {P=f,Q=f, R=t, S=t}, the premises P → Q, R ˄ S, are both true, but ¬P → (Q ˄ R) is false.

share|improve this answer

edited 53 mins ago

answered 1 hour ago

Graham KempGraham Kemp

1,10919

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No, there's no typo. I imagined that this proof wasn't valid simply because I coulnd't conclude (NOT)P from anywhere. So, when I replace the letters with cube(a) , cube(b), why is it showing, in the world, that the proof is valid? Should I choose names that I know will prove it's invalid?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  

add a comment | 

2

P → Q, R ˄ S |- ¬P → (Q ˄ R) is not syntactically valid.

An assumption of ¬P will not allow you to eliminate the conditional in P → Q to derive Q .

Is there a typo?

P → Q, R ˄ S |- P → (Q ˄ R) is valid.

 1|  P → Q

 2|_ R ˄ S

 3|  R               ˄E, 2

 4|  |_ P

 5|  |  Q            →E, 1,4

 6|  |  Q ˄ R        ˄I, 3,5

 7|  P → (Q ˄ R)    →I, 4-6

---

For an argument to be semantically valid, the conclusion must be demonstrably true in all interpretations where the premises are -- it is not enough to find just one. A proof is semantically invalid when the exists some interpretation where all the premises are true, but the conclusion is false. One such counter example is enough to disprove an argument.

In the interpretation of {P=f,Q=f, R=t, S=t}, the premises P → Q, R ˄ S, are both true, but ¬P → (Q ˄ R) is false.

share|improve this answer

edited 53 mins ago

answered 1 hour ago

Graham KempGraham Kemp

1,10919

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No, there's no typo. I imagined that this proof wasn't valid simply because I coulnd't conclude (NOT)P from anywhere. So, when I replace the letters with cube(a) , cube(b), why is it showing, in the world, that the proof is valid? Should I choose names that I know will prove it's invalid?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  

add a comment | 

P → Q, R ˄ S |- ¬P → (Q ˄ R) is not syntactically valid.

An assumption of ¬P will not allow you to eliminate the conditional in P → Q to derive Q .

Is there a typo?

P → Q, R ˄ S |- P → (Q ˄ R) is valid.

 1|  P → Q

 2|_ R ˄ S

 3|  R               ˄E, 2

 4|  |_ P

 5|  |  Q            →E, 1,4

 6|  |  Q ˄ R        ˄I, 3,5

 7|  P → (Q ˄ R)    →I, 4-6

---

For an argument to be semantically valid, the conclusion must be demonstrably true in all interpretations where the premises are -- it is not enough to find just one. A proof is semantically invalid when the exists some interpretation where all the premises are true, but the conclusion is false. One such counter example is enough to disprove an argument.

In the interpretation of {P=f,Q=f, R=t, S=t}, the premises P → Q, R ˄ S, are both true, but ¬P → (Q ˄ R) is false.

share|improve this answer

edited 53 mins ago

answered 1 hour ago

Graham KempGraham Kemp

1,10919

 1|  P → Q

 2|_ R ˄ S

 3|  R               ˄E, 2

 4|  |_ P

 5|  |  Q            →E, 1,4

 6|  |  Q ˄ R        ˄I, 3,5

 7|  P → (Q ˄ R)    →I, 4-6

share|improve this answer

edited 53 mins ago

answered 1 hour ago

Graham KempGraham Kemp

1,10919

answered 1 hour ago

Graham KempGraham Kemp

1,10919

answered 1 hour ago

Graham KempGraham Kemp

1,10919

1

By using a truth table generator one can show that one of the set of valuations for the sentence letters leads to the result being false.

To see this, use this input ((P=>Q)&&(R&&S))=>(~P=>(Q&&R)) in the Stanford Truth Table Tool. When P and Q are false and R and S are true then the conditional with the conjunction of the premises as antecedent and the goal as consequent is false.

This would mean that one could not derive the result.

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That makes sense. So the reason that the proof is invalid, is because out of all the results, there's only 1 that is false? In a valid proof, they should all be true?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That's right. If there were only T values then the statement would be a tautology based on the truth table and a derivation or proof could be found that would be valid.
  
  – Frank Hubeny
  1 hour ago
  

  
  
  
  

add a comment | 

1

By using a truth table generator one can show that one of the set of valuations for the sentence letters leads to the result being false.

To see this, use this input ((P=>Q)&&(R&&S))=>(~P=>(Q&&R)) in the Stanford Truth Table Tool. When P and Q are false and R and S are true then the conditional with the conjunction of the premises as antecedent and the goal as consequent is false.

This would mean that one could not derive the result.

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That makes sense. So the reason that the proof is invalid, is because out of all the results, there's only 1 that is false? In a valid proof, they should all be true?
  
  – fitch-is-killing-me
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That's right. If there were only T values then the statement would be a tautology based on the truth table and a derivation or proof could be found that would be valid.
  
  – Frank Hubeny
  1 hour ago
  

  
  
  
  

add a comment | 

By using a truth table generator one can show that one of the set of valuations for the sentence letters leads to the result being false.

To see this, use this input ((P=>Q)&&(R&&S))=>(~P=>(Q&&R)) in the Stanford Truth Table Tool. When P and Q are false and R and S are true then the conditional with the conjunction of the premises as antecedent and the goal as consequent is false.

This would mean that one could not derive the result.

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

share|improve this answer

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559

answered 1 hour ago

Frank HubenyFrank Hubeny

11k51559