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June 1st, 2008
10:34 pm - More logic
So I was looking around on the internet at fallacies, and found an example of the Relativist Fallacy which seems appropriate to the example I used for my last post on logic.
Jill: "Look at this, Bill. I read that people who do not get enough exercise tend to be unhealthy."
Bill: "That may be true for you, but it is not true for me."
But is this really a Relativist Fallacy? Disregarding what we know about the world (as what we're talking about here is the logical structure of the argument, not the facts behind the argument), it seems pretty poorly worded to hold up as an example. "People who do not get enough exercise tend to be unhealthy" is the premise being refuted here. To say that that is true for one person and not for another is, of course, not merely fallacious but downright silly, since the premise makes a categorical statement. However, in the wording "tend to", there is an implication that not everybody who does not exercise is unhealthy, but most are. So wouldn't, then, the response "that's not true for me" be read, not as a statement that categorical premises can somehow not hold for certain people, but as a statement that the speaker is a member of the minority implied by "tend to" in the original premise?
My normal approach kind of fails here, since I don't have the logical tools to represent 'most' or 'tend to', and even if I did, the relationship between the English sentences and their symbolic counterparts is ambiguous at best.
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May 28th, 2008
11:43 pm - Does anyone know what the fallacy here is called?
Suppose you have a formula ¬∀x¬(fat(x)→healthy(x)).
Now someone tries to disprove it by showing that ¬∀x(fat(x)→healthy(x)).
This is clearly a False Dilemma fallacy, because it ignores the fact that ¬∀x¬(fat(x)→healthy(x)) means ∃x(fat(x)→healthy(x)), NOT ∀x(fat(x)→healthy(x)), whereas ¬∀x(fat(x)→healthy(x)) means ∃x¬(fat(x)→healthy(x)), and provided there is more than one person in the world, ∃x(fat(x)→healthy(x))∧∃x¬(fat(x)→healthy(x)) is not a contradiction.
However it seems to me that it should be a special case of the False Dilemma, because the dilemma is not merely in asserting that there are only two possibilities when there are actually more, but in fact that false dilemma comes from an ignorance of the relationship between universality and existentiality. So if it is a special case, does it have a special name?
Anyone?
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September 18th, 2007
09:56 am - Fundie logic. No, really this time.
More fundie gold this morning. I can't remember the exact quote, but it went something like this:-
"De Morgan's Law spells death for science. Consider the following propositions: P) 1+1=3 and Q) 2+2=5. We know these propositions are false, but because of De Morgan's Law, if we take (P∧Q) they are both true! Take that, science!"
Er, what? De Morgan's Law says nothing of the sort! What you are saying there is that:-
(P∧Q)⇔¬(P)∧¬(Q)
Where ∧ is a logical conjunction symbol (i.e. AND), ¬ is a logical negation symbol (i.e. NOT) and ⇔ is a logical equivalence symbol (i.e. the thing on the left is the same as the thing on the right). That is: The conjunction of P and Q is logically equivalent to the conjunction of the negation of P and the negation of Q.| Truth table for (P∧Q)⇔¬(P)∧¬(Q) |
| P | Q | ¬P | ¬Q | P∧Q | ¬(P)∧¬(Q) |
| T | T | F | F | T | F |
| T | F | F | T | F | F |
| F | T | T | F | F | F |
| F | F | T | T | F | T |
WRONG!
What De Morgan's Law says is:-
¬(P∧Q)⇔¬(P)∨¬(Q)
(The negation of the conjunction of P and Q is logically equivalent to the disjunction (i.e. OR) of the negation of P and the negation of Q.)| Truth table for ¬(P∧Q)⇔¬(P)∨¬(Q) |
| P | Q | ¬P | ¬Q | (P∧Q) | ¬(P∧Q) | ¬(P)∨¬(Q) |
| T | T | F | F | T | F | F |
| T | F | F | T | F | T | T |
| F | T | T | F | F | T | T |
| F | F | T | T | F | T | T |
Right!
¬(P∨Q)⇔¬(P)∧¬(Q)
(The negation of the disjunction of P and Q is logically equivalent to the conjunction of the negation of P and the negation of Q.)
| Truth table for ¬(P∨Q)⇔¬(P)∧¬(Q) |
| P | Q | ¬P | ¬Q | (P∨Q) | ¬(P∨Q) | ¬(P)∧¬(Q) |
| T | T | F | F | T | F | F |
| T | F | F | T | T | F | F |
| F | T | T | F | T | F | F |
| F | F | T | T | F | T | T |
Right!
Which is almost the opposite of what you said. But don't get too heartened that I use the word 'almost'. By that word, I mean it actually has nothing whatsoever to do with what you said.
The conjunction of P and Q, by the way, is funnily enough logically equivalent to, um, the conjunction of P and Q, as expressed by: (P∧Q)⇔(P)∧(Q). If you couldn't guess.
But then, I guess this fundie will never read my livejournal, and the rest of you don't need me to tell you fundies don't have the best grasp of logic, right?
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