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## Comments

If cats like dogs then my terrier will not bite the kitten unless it is angry.This gives rise to so many more worlds than a more simple conditional.(CD)----->(/A----->/B)Yes, there's more room for error if one translates the phrases incorrectly. The best way to learn not to translate incorrectly would be to check out the lesson on embedded conditionals https://7sage.com/lesson/mastery-embedded-conditional/When I see a conditional phrase in a statement on the LSAT, I translate each statement within each section of the conditional phrase and then apply the conditional phrase indicators (for the larger sentence) to those phrases.

The contrapositive of your schema:

(~M & ~B) -> ~P

But your vernacular feels more like a biconditional to me:

P->(M<->~B)

'Reason being it seems to assert that, indeed, Mike WILL be there in the abscence of the birds, in addition to the birds guaranteeing Mike's abscence, so the abscence of one is necessary and sufficient for the abscence of the other.

In that case, the contrapositive:

[(M & or (~M & ~B)] -> ~P

- which has a long form of the negated biconditional:

(M<->B) -> ~P

Finally, just want to remark on the potential for scope ambiguity in the vernacular. You could read it to mean:

(P->M) <-> ~B

In that case, the birds being at the park would guarantee your own presence there (without seeing Mike). 'Very influential birds, at any rate.

But that comma really clears it up by placing the "unless" grammatically within the consequent of the leading conditional.

The contrapositiveof P --->(~M--->B)is ~B---> M or ~PWhy did I get that? I think the best way to start is to first

simplify the statement P --->(~M--->B) before figuring out the contrapositive. To do that,take the embedded phrase (the one in the parenthesis) and move it outside of the parenthesis by changing the first arrow to an "and"and keeping the other terms the same. So itchanges from P---> (~M--->B)toP and ~M --->B(or P and ~B ---> M). Which means: if I go to the park and do not see Mike, then the birds are there.How/why did we do that? What's happening here is

we now have two sufficient terms that should occur to give us the necessary condition(which isgoing to the parkandnot seeing mike)being sufficient to. Below I provided the logistic proof for why/how, using the negate sufficient "or" rule with the arrows you can arrive at the simplified answer that I did. Use the "or" arrow rule (which is negating the sufficient term that you remove the arrow for along each step in the proof starting with the original statement) the following is just the proof and a step by step application of the arrow to or translation rule.the birds being thereP --->(~M--->B)~P or (~M--->B)

~P or (M or

[~P or M] or B

not[~P or M] --->B *apply De Morgan's law (basically two negations cancel each other out)

P and ~M ---->BThe contrapositive of the last logical statement (our simplified statement) is ~B---> M or ~P

(don't forget that you switch "and" with "or" when getting the contrapositive).

Hmm @Q.E.D Doesn't P--->(~M-->B) mean going to the park is sufficient for not seeing Mike, while not seeing Mike is sufficient for birds being there? Or stated another way P--->(~B--->M) : going to the park is sufficient for the birds not being there, which is sufficient for seeing Mike? The relationship of the two terms B and M are such that they must exist in an oppositely negated relationship to each other. B and M are both not going to be positive or both be negative in the same statement because the term "unless" in the provided sentence requires that we negate one of those terms and make it a sufficient condition for the other.

Cool stuff you're putting out there. I wish more people had interests like yours.

'Just gonna go ahead and hit these prompts. First, I couldn't agree more with your thoughts on the biconditional: And in response to these comments... I read (1) to mean P is sufficient for the sufficiency of ~M for B. P guarantees you can't have ~M without B, though possibly M is true - in which case (on this schema), either B or ~B. Here's where the familiar LG-type chains, though fantastic, don't capture critical scope ambiguities in cases like this. In the games, you often see things like P->~M->B, where every item is sufficient for the next individual item down the line, as in the quoted interpretation. In sentence logic, however, you often have to explicitly set the scope of each conditional to apply to a connective (e.g. ->) rather than a propositional atom, as in (1), where the necessary condition doesn't itself guarantee the truth of anything.

Here is some scope syntax totally lost on the straight chain:

P--->(~M-->B)

(P--->~M)-->B

In that vein, I would add about the proof above that the endpoint has a scope ambiguity.

'P and ~M ---->B'

could be

(2) (P and ~M) ---->B or

(3) P and (~M ---->B),

...where I'd advocate for (2).

Awesome proof, btw.

Best,

QED

I think a lot of the confusion here stems from the fact that it's unclear whether "unless" expresses a conditional or a biconditional. For LSAT purposes, these are treated as straightforward conditionals, but in ordinary language, they can behave like biconditionals. This isn't a settled issue among logicians, but I think the trend is towards interpreting "unless" as a conditional and explaining instances that appear to be biconditionals as implicatures.

@nessa.k13.0

Remember that '(A -> ' is equivalent to '(~A or ', so the '~(A -> ' is the same as '~(~A or ', which begets '(A and ~B)'. Plugging '~M' and 'B' in for those schematics gets us from '~(~M -> ' to '~(~~M or ', hence '~(M or ' and '(~M and ~B)'.

It also makes sense to remember that a conditional is only false when the antecedent is true and the consequent is false, as in '(A and ~B)' above. Incidentally, you'll never see a bald negation of a conditional connective on the LSAT, e.g. one that translates as '~(A -> ', because the material conditional we're negating in this thread doesn't quite capture natural language conditionals. That is, an LSAT prob might set us up to infer '~(A -> ' from '(A and ~B)', but never the other way around. Consider this example:

(1) It's not true that if Jones is arrested then he'll confess.

You would schematize it as '~(A -> ', which translates to '(A and ~B)'. Tense aside, classical logic would have us conclude that John was in fact arrested and did not confess, whereas we don't naturally see any commitment to these apparent facts in English. Turns out only non-classical logics have the expressive power to approximate most natural language conditionals. From a formal point of view, it seems the LSAC test writers have some uncomfortable limitations in how they can write questions.

Excuse me. I get excited about logic. On that note, I need to know how quinnxzhang knows that Do you follow the literature on symbolic logic, @quinnxzhang? Your awareness of correct use-mention notation, as in has analytic philosophy written all over it.

Anyway, good luck on the LSAT to all. 'Fun thread.

http://fitelson.org/probability/macfarlane.pdf

It looks solid, man. I haven't seen a lot of literature on the demarcation problem in logic. I think I'll dig into this.

Looks like it covers 2nd order logic. I'm aware that Quine considered it "set theory is sheep's clothing" and bumped it across that demarcation line out of logic, but I never researched the controversy.

Nice crossing paths.

truth function'-->'. The conditional takes in the truth values of the antecedent '~M' and consequent 'B', and then it spits out either true or false.Remember the definition

of the material conditional (A --> C):

True when A is false and C is true;

True when A is false and C is false;

True when A is true and C is true;

False when A is true and C is false.So plug in '~M' and 'B' for A and C in the definition. To negate

'(~M -> ' is to say it's false, and there's only one condition where the '-->' is false, namely when '~M' is true and 'B' is false, hence '~(~M --> ' is (~M and ~B)'.

A bit painstaking but very clear, I hope.

As you said, '~(~M)' is just 'M', but notice the scope of the negation only encompasses '~M'. In the example you cited, '~(~M -> ', the negation outside those parentheses operates on everything inside the parentheses. So it's actually the conditional relationship '-->' between '~M' and 'B' that's being negated, not necessarily '~M'. Does that make sense?