## Logical Fallacy Game

Philosophical, mathematical and computational logic, linguistics, formal argument, game theory, fallacies, paradoxes, puzzles and other related issues.

Bump. Let's keep this thing going. We're still on:

Therefore, there are some bad habits that everyone has.

Sparky
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Last bump. Anyone?

Sparky
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Can't tell you the exact term, but it can't be logical, as it doesn't imply that all people have the same bad habit at least once.
dotty
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dotty,

Yes. It's called the Quantifier-Shift Fallacy.

Sparky
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### Fallacy

Let's try this one:

In Shakespeare's "Merchant of Venice" Shylock claims a pound of flesh near to Antonio's heart, because A. couldn't pay his debt on time. But the courts deny him his right by forbidding him spill Antonio's blood. Shylock can't cut his pound of flesh without drawing blood, so he can't claim his right.

There is a fallacy in this situation. What is it called?
dotty
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I think I know what it is, but I will wait to let someone else have a stab at it.

Sparky
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In Shakespeare's "Merchant of Venice" Shylock claims a pound of flesh near to Antonio's heart, because A. couldn't pay his debt on time. But the courts deny him his right by forbidding him spill Antonio's blood. Shylock can't cut his pound of flesh without drawing blood, so he can't claim his right.

Is it equivocation? The phrase "near to Antonio's heart" is ambiguous because it could have literal meaning (as apparently interpreted by the courts), or it could have a figurative meaning, which could simply mean "valued emotionally."

Sparky
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### Fallacy

You're right. It's an equivocation, similar to Loki's Wager.

dotty
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Jesus was either a liar, a lunatic, or Lord.
Jesus was neither a liar nor a lunatic.
Therefore, Jesus is Lord.

Sparky
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Fallacy of the false Dilemma: Jesus was liar, lord, or lunatic, does not exhaust all possibilities.

cougar
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Sparky
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This one is pretty easy, generally, why does this fail?

For all true statements, there is some object that they describe.
So, there must be some object which all true statements describe.

cougar
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For all true statements, there is some object that they describe.
So, there must be some object which all true statements describe.

Quantifier-Shift Fallacy?

Sparky
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Yep, let "Describe" be a two place predicate $\mathcal{D}$, for $S = \{ \phi | \phi \text{ is a True statement} \}, O = \{ \phi | \phi \text{ is an Object }\}, S \times O \rightarrow \mathcal{D}$. Then, our derivation is the following, $(\forall x \in S. \exists y \in O. \mathcal{D}(x,y)) \rightarrow (\exists x \in O. \forall y \in S. \mathcal{D}(x,y))$, which is not true in general.

To see why, lets pick an arbitrary $x \in S$. Assume that a particular object, call it $a$ holds for $\mathcal{D}(x,y)$. Now we would have to show that $\mathcal{D}(x,y)$ holds for that particular $x$ for for any $y \in O$. But, since $x$ was arbitrary, we don't know enough to do this generally, we just know it works in one particular case, not all.

cougar
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cougar wrote:Yep, let "Describe" be a two place predicate $\mathcal{D}$, for $S = \{ \phi | \phi \text{ is a True statement} \}, O = \{ \phi | \phi \text{ is an Object }\}, S \times O \rightarrow \mathcal{D}$. Then, our derivation is the following, $(\forall x \in S. \exists y \in O. \mathcal{D}(x,y)) \rightarrow (\exists x \in O. \forall y \in S. \mathcal{D}(x,y))$, which is not true in general.

To see why, lets pick an arbitrary $x \in S$. Assume that a particular object, call it $a$ holds for $\mathcal{D}(x,y)$. Now we would have to show that $\mathcal{D}(x,y)$ holds for that particular $x$ for for any $y \in O$. But, since $x$ was arbitrary, we don't know enough to do this generally, we just know it works in one particular case, not all.

I don't think you are really done, cougar.

It'd be better to give a model where the antecedent is true but the predicate is false, to really show that it is invalid. An easy one is S={a,b} O={c,d} D={(a,c),(b,d)}; there are probably easier ones.

-------

Here's one I came up with the other day, from Provability Logic.

As you may all know, Bew('P') is Gödel's notation for the statement "There exists a proof of proposition P" where 'P' is the numerical representation of P. Furthermore, there is a famous theorem called Löb's Theorem, that says Bew('Bew('P')->P')->Bew('P').

Now, suppose ~Bew('A') for some A.
Then clearly, we have Bew('A')->A, because the antecedent is false.
Thus, we know that Bew('Bew('A')->A'), so I may use modus ponens to deduce Bew('A')
So I know by Reductio ad Absurdem that Bew('A'), for any wff A. In other words, any statement is provable.

Note to Sparky: I don't think this fits into a cookie-cutter fallacy type...

xcthulhu
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cloudy-a wrote:The current logical-fallacy-containing statement is:
"Good and evil are constantly battling it out for control of the universe"
sparky wrote:This is a tough one. Is it the "Figure of Speech" fallacy?
The phrase figuratively gives "good" and "evil" human-like qualities by implying that they can battle, which may cause confusion.
cloudy-a wrote:Yep Sparky. I learned it as reification fallacy, but I think it is the same thing (?)
Anyhow, you're up. :)

isn't re-ification to "thing"-ify, while to give something human-like qualities is to "person"-ify?
sophist413
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xcthulhu wrote:
cougar wrote:Yep, let "Describe" be a two place predicate $\mathcal{D}$, for $S = \{ \phi | \phi \text{ is a True statement} \}, O = \{ \phi | \phi \text{ is an Object }\}, S \times O \rightarrow \mathcal{D}$. Then, our derivation is the following, $(\forall x \in S. \exists y \in O. \mathcal{D}(x,y)) \rightarrow (\exists x \in O. \forall y \in S. \mathcal{D}(x,y))$, which is not true in general.

To see why, lets pick an arbitrary $x \in S$. Assume that a particular object, call it $a$ holds for $\mathcal{D}(x,y)$. Now we would have to show that $\mathcal{D}(x,y)$ holds for that particular $x$ for for any $y \in O$. But, since $x$ was arbitrary, we don't know enough to do this generally, we just know it works in one particular case, not all.

I don't think you are really done, cougar.

It'd be better to give a model where the antecedent is true but the predicate is false, to really show that it is invalid. An easy one is S={a,b} O={c,d} D={(a,c),(b,d)}; there are probably easier ones.

-------

Here's one I came up with the other day, from Provability Logic.

As you may all know, Bew('P') is Gödel's notation for the statement "There exists a proof of proposition P" where 'P' is the numerical representation of P. Furthermore, there is a famous theorem called Löb's Theorem, that says Bew('Bew('P')->P')->Bew('P').

Now, suppose ~Bew('A') for some A.
Then clearly, we have Bew('A')->A, because the antecedent is false.
Thus, we know that Bew('Bew('A')->A'), so I may use modus ponens to deduce Bew('A')
So I know by Reductio ad Absurdem that Bew('A'), for any wff A. In other words, any statement is provable.

Note to Sparky: I don't think this fits into a cookie-cutter fallacy type...

I think I've seen this one before. It's called "argumentum ad no comprehendum"

... otherwise know as the "baffle you with my brilliance" fallacy. ;-)

j/k, care to explain where the fallacy lies?

Sparky
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Sparky wrote:
xcthulhu wrote:As you may all know, Bew('P') is Gödel's notation for the statement "There exists a proof of proposition P" where 'P' is the numerical representation of P. Furthermore, there is a famous theorem called Löb's Theorem, that says Bew('Bew('P')->P')->Bew('P').

Now, suppose ~Bew('A') for some A.
Then clearly, we have Bew('A')->A, because the antecedent is false.
Thus, we know that Bew('Bew('A')->A'), so I may use modus ponens to deduce Bew('A')
So I know by Reductio ad Absurdem that Bew('A'), for any wff A. In other words, any statement is provable.

Note to Sparky: I don't think this fits into a cookie-cutter fallacy type...

I think I've seen this one before. It's called "argumentum ad no comprehendum"

... otherwise know as the "baffle you with my brilliance" fallacy. ;-)

j/k, care to explain where the fallacy lies?

Sometimes I've heard it as argument by intimidation. It's not a nice thing to do, I'm sorry if I came across that way :(

The short of it is that provability logic is an example of an intuitionistic logic, which doesn't have the rule Reductio Ad Absurdem; so the last step is invalid.

Pretty tricky, right? My logic professor Henry Kyburg informed me of this fallacy when I committed it in a similar derivation a while back.

Anyway, someone else's turn...

xcthulhu
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### Re: Logical Fallacy Game

Here's one:

Michael Behe is a tenured professor at Lehigh university and he believes in intelligent design.
Therefore, intelligent design is true.

linford86
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### Re: Logical Fallacy Game

linford86 wrote:Here's one:

Michael Behe is a tenured professor at Lehigh university and he believes in intelligent design.
Therefore, intelligent design is true.

Argument from authority?

kidjan
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### Re: Logical Fallacy Game

kidjan wrote:
linford86 wrote:Here's one:

Michael Behe is a tenured professor at Lehigh university and he believes in intelligent design.
Therefore, intelligent design is true.

Argument from authority?

Yep, otherwise known as argumentum ad verecundiam. Who's next?

linford86
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### Re: Logical Fallacy Game

I was just looking that one up the other day as someone accused me of committing it.

Deftil
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### Re: Logical Fallacy Game

Here's one.

"I think therefore I am".

Some people believe this is a fallacy, others believe it true. I suppose this thread will offer some opinion and thought on this, figured I'd post it to see.

Timothy
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### Re: Logical Fallacy Game

No rice is snow
No snow is hot
Thus, no rice is hot
philip8
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### Re: Logical Fallacy Game

philip8 wrote:How about . . .

No rice is snow
No snow is hot
Thus, no rice is hot

1. $\forall x(R(x) \rightarrow \neg S(x))$
2. $\forall x(S(x) \rightarrow \neg H(x))$

$R(a) \rightarrow \neg S(a)$ from universal elimination of (1)
Assume $R(a)$ for conditional proof

$\neg S(a)$ from modus ponens
$S(a) \rightarrow \neg H(a)$ from universal elimination of (2)

Nowhere to go from there. I guess the fallacy would be denying the antecedent?
Hylas
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### Re: Logical Fallacy Game

Hylas wrote:Nowhere to go from there. I guess the fallacy would be denying the antecedent?

Fallacy files lists this as "Exclusive Premisses"
http://www.fallacyfiles.org/exclprem.html

On the other hand, since you are symbolically inclined, we can always tell when some syllogistic inference pattern, or for that matter any elementary pattern in propositional logic is invalid, simply by making a counter-model. If you are so interested, I could show you how to use the computer program mace to do this...

xcthulhu
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### Re: Logical Fallacy Game

Can we post informal fallacies? If so, here's one:

Marijuana is completely natural (it's a plant!)
Anything that is completely natural is good (or healthy.)
Therefore, marijuana is good (or healthy.)

linford86
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### Re: Logical Fallacy Game

linford86 wrote:Can we post informal fallacies? If so, here's one:

Marijuana is completely natural (it's a plant!)
Anything that is completely natural is good (or healthy.)
Therefore, marijuana is good (or healthy.)

1) Marijuana is completely natural (it's a plant!)

Not entirely true since a lot of marijuana is cultivated by humans which might use artificial fertilizers, controlled irrigation and probably includes selection of “stronger” lineages ( more cannabinoids content ). Therefore part of it is not entirely “natural” ( if by natural we mean something that did not suffer human technological manipulation ).

2) Anything that is completely natural is good (or healthy.)

This is largely false. For instance it is known that many plants contain toxins or are even poisonous for human consumption. Some of them have even carcinogenic agents. Most of contagious diseases caused by pathogens and parasites are also natural and yet they are mostly bad to our health. Etc..

3) Therefore, marijuana is good (or healthy.)

This fallacy contains largely false and grossly inaccurate premises which do not guarantee the veracity of the conclusion. And yet the assertion of the conclusion could still be correct or not.

It is even more complex. The conclusion could be false or correct depending of external conditions. For instance, if marijuana is used for medicinal purpose then its benefit could outweigh its damaging side effects. If it is used heavily and frequently as an addiction the result could be the opposite. Therefore something could be partially true and partially false depending of external ( contextual ) conditions.

There are even additional relevant, but a bit hair-splitting, arguments to be made in this case but I will stop here.
Paris
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### Re: Logical Fallacy Game

linford86 wrote:Can we post informal fallacies? If so, here's one:

Marijuana is completely natural (it's a plant!)
Anything that is completely natural is good (or healthy.)
Therefore, marijuana is good (or healthy.)

Fallacy of Accident?

http://www.fallacyfiles.org/accident.html

mtbturtle
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### Re: Logical Fallacy Game

I removed my examples of fallacies because I intend to discuss them in a more versatile and deep way than that imposed by the rules of this game.

I will post them in the new topic “Detect the Fallacies”.
Paris
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