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)
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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)
It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.
logic deduction
edited 8 hours ago
Frank Hubeny
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It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.
logic deduction
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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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
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.
It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.
logic deduction
logic deduction
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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edited 8 hours ago
Frank Hubeny
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asked 9 hours ago
zzzzzzzz
91
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zzzz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 8 hours ago
Frank Hubeny
11.7k51564
edited 8 hours ago
Frank Hubeny
11.7k51564
edited 8 hours ago
Frank Hubeny
11.7k51564
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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asked 9 hours ago
zzzzzzzz
91
asked 9 hours ago
zzzzzzzz
91
91
New contributor
zzzz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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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 |
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
I made an edit which you may roll back or continue editing. Welcome!
– Frank Hubeny
8 hours ago
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
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 |
2 Answers
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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
answered 8 hours ago
Frank HubenyFrank Hubeny
11.7k51564
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).
answered 5 hours ago
EliranEliran
4,52231435
add a comment |
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2 Answers
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2 Answers
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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
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
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
answered 8 hours ago
Frank HubenyFrank Hubeny
11.7k51564
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
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
answered 8 hours ago
Frank HubenyFrank Hubeny
11.7k51564
11.7k51564
add a comment |
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).
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).
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).
answered 5 hours ago
EliranEliran
4,52231435
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).
answered 5 hours ago
EliranEliran
4,52231435
answered 5 hours ago
EliranEliran
4,52231435
answered 5 hours ago
EliranEliran
4,52231435
answered 5 hours ago
EliranEliran
4,52231435
4,52231435
add a comment |
add a comment |
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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