My criticism of the rants (those are not reviews) is independent of the books. What she said about logic is stupid, no matter what is in the books.

I looked only at the tables of contents of the books. The books seem to be meant for general students and to give a first overview of basic logic. My guess is that they are fine for that purpose. I would suggest studying them if you have nothing else. Kalish/Montague/Mar covers much of the same ground, but it's the best introduction I've seen, though probably it's more demanding than the other two books.

I started studying logic on my own. My first logic textbook didn't even have symbolic logic. Mostly about informal fallacies, a bit about syllogisms, and the difference between deductive logic and inductive logic. It was okay I guess. Then I worked through an introductory book on symbolic logic, and I learned a lot. Then I got Kalish/Montague (Mar was not a co-author of that edition). It covered basically the same ground as my first symbolic logic book, but I saw that it did a vastly better job of it. But I will say that I just don't know whether I would have done as well with Kalish/Montague if I hadn't previously read that other symbolic logic book that gave me some good chops with symbolization and symbolic deduction.

"Good sense about trivialities is better than nonsense about things that matter." -- Quote on a math professor's door that I saw once. — fishfry

That's pretty good. I like it.

But the quoted parts are what I felt was good points. — Corvus

They're not. They reveal fundamental misunderstanding, confusions, and ignorance of the writer.

I read both her posts. They're ignorant diatribes.

he reviewer didn't sound like a newbie (she has many Logic books, and read them all) — Corvus

Based on the quote you provided, she seems not to understand what she read, thus remaining ignorant.

there were parts that resonated with my feelings about the books. — Corvus

You just quoted her about the ill-effects of emotion in arguments. Your feelings about the books don't make her arguments about them sound.

I see your point. But I haven't been in disagreement.

I don't dispute the author's argument about the modus ponens argument.

My point is to be careful not to take his example in the form he literally gave it.

P -> Q
P
Therefore [modal]Q.

Rather that his analysis can be stated along the lines of:

If modus ponens is valid, then if we believe the premises, then we believe the conclusion (not always in fact - people err - but in principle). And the premises are believed but the conclusion is not. — TonesInDeepFreeze

This does not prove the invalidity of modus ponens. Rather it shows that modus ponens may fail our expectations of belief. And it does seem to me to be a genuine puzzle.

When logic is used in the debates, the debaters might get a false sense of security that they might arrive at true conclusions because they are using logical methods. But in many cases, it is not the case. Because logic can hide the traps. — Corvus

Sure, people err, and abuse even simple logic. And logic is often not simple.

Logic by Wilfrid Hodges
Introduction to Logic by Gensler — Corvus

Thanks, I'll look at them out if I see them somewhere.

Quotes below are from a third party's comments on a book:

"This book, like other pseudo-logic texts of the type, does inform us that logical arguments require true premises."

'pseudo-logic' is mere and false characterization.

Good for those books. Logical arguments do not require true premises.

"technical logical rules designed to insure that we are able to spot obvious logical contradiction in an argument"

Wrong. The rules are to prevent non sequiturs, and to disallow inferring contradictions from consistent premises.

The writer of the review doesn't understand logic.

"I know of no one who will believe an argument which they know is derived from untrue premises, or which contains obvious contradiction."

The writer fails to understand the difference between evaluation of the validity of an argument and the truth value of the conclusion.

"The real problem is that via appeals to authority and emotion we tend to accept premises as true which are not supported by evidence or which are deceptively incomplete."

Appeals to emotion and authority are informal fallacies. Such fallacies are worth discussing, but they are not in the field of formal logic.

"Gensler and his ilk teach us what we we already know naturally."

Yeah, I don't think so. The writer doesn't even know the difference between inferential validity and factual truth and falsity.

You are quoting from someone who is ignorant.

The conclusion is not valid. The conclusion is contingent.

The modus ponens argument

(R v A) -> (~R -> A)
R v A
therefore ~R -> A

is valid.

But the argument is not of that form. It's of this form:

(R v A) -> (~R -> A)
R v A
therefore [modal operator] ~R -> A

But there is still a puzzle, as I mentioned.

the argument as stated can't be interpreted as a modus ponens — fdrake

The argument as stated is not modus ponens. It injects a modal operator in front of the conclusion.

But there is still a puzzle:

If modus ponens is valid, then if we believe the premises, then we believe the conclusion (not always in fact - people err - but in principle). And the premises are believed but the conclusion is not. So, still, there's a puzzle. — TonesInDeepFreeze

Maybe I'll get time to analyze the rest of your post.

Of course, but that doesn't address the puzzle.

"therefore we have deductive reason to assert" — bongo fury

That still breaks the form of modus ponens.

'we have deductive reasons to assert is' is intensional.

If it were merely a flourishing touch, then we could delete it, but if we delete it then then the puzzle fizzles in the form it's given, as its form is not modus ponens if it injects a modal operator.

If P then Q
P
Therefore Q

is modus ponens.

If P then Q
P
Therefore [modal operator] Q

is not modus ponens.

extensionalism — bongo fury

On the contrary, it introduces intensionality.

the enemy of logic is often, the logic itself — Corvus

How so?

I have a couple of basic logic books, but they seem not great. — Corvus

Which books are those?

What is your mathematical definition of 'infinites'?

Unlike the odd numbers, there is no set of all infinite sets nor of all infintie cardinals. So what are the infinities?

We could define "x is an infinity iff x is infinite'. But then 'is an infinity' is a predicate and doesn't stand for a quantity.

A cardinality is a quantity. But 'is infinite' is an adjective.

It's not a merely pedantic distinction. Ignoring the distinction causes real confusions about sets and set theory. Ignoring the distinction is typical of cranks (not you) who know nothing about set theory but try to refute it with incoherent arguments that conflate 'is infinite' as if 'infinity' is a noun.

one infinity can be bigger than another infinity — DingoJones

One infinite set may be larger than another infinite set.

Infinite is a quality, not a quantity.
— Possibility
Which is just wrong. — Banno

It's right. The quantities are the particular cardinalities. No cardinality itself is 'infinity'. Rather, each infinite cardinality has the property of being infinite.

'is infinite' is a predicate. a set is infinite iff it is not finite.

'infinity' as a name occurs as sometimes, such as points in the extended real system. But that is different from the sense 'is infinite'. And such points are not required themselves to be infinite sets, though in some treatments they are.

'to infinity' in such statements as regards limits is not a separable term but abides only in the structure of a larger notation.

This post is not addressed to any specific person.

[...] "ExAy yex" has models.
— GrandMinnow

But without further qualification, those models are in no way sets. — fishfry

I am GrandMinnow. I hadn't gotten around to answering the above.

Here is a model of "ExAy yex":

<{0} {<'e' {<0 0>}>}>

And its domain is a set, and the model itself is a set.

If one is familiar with mathematical logic, then one should recognize that is a model of "ExAy yex". It is quite trivial. But, to me, the very fact that it is so trivial is interesting.

But if one is not familiar with mathematical logic, then one would not understand the above. I am going to explain it exactly as I can and with as much detail as is feasible (even if painstakingly pedantic) in the context of posting.

(1) My original point was that "ExAy yex" is consistent but that it is not consistent with the axiom schema of separation. Note that "ExAy yex" is not consistent with this instance of the axiom schema of separation: AzEyAx(xey <-> (xez & ~xex)).

I mentioned that to point out that ruling out a universal set is not just a matter of looking alone at the notion of a universal set but rather that the notion of a universal set is not consistent with the notion of comprehension.

Then I said that "ExAy yex" has models. Note that if a sentence has a model, then the sentence is consistent.

(2)

(a) Df: a set of formulas T is consistent iff T does not prove a contradiction.

(b) Df: a formula P is consistent iff {P} is consistent.

If I recall correctly, (b) is fairly standard, but perhaps some people prefer to provide only (a).

I mention that only to ward against quibbles that originally I said:

"ExAy yex" is consistent

instead of

{"ExAy yex"} is consistent.

(3) Let 'N' stand for the set of natural numbers.

Df: a first order language L is determined by a signature <F B t> such that the intersection of F and B is empty, and F is the set of function symbols for L, and B is the set of relation symbols for L, and t is a function from FuB into N. For any function or relation symbol s, t(s) is called "the arity of s". We will leave 'first order' tacit in the rest of the post.

Df: L is a language for (or 'of') a set of formulas T iff L is a language that has at least all the function symbols and relation symbols that occur in T, and with the same arity they have in the formulas. Of course, this makes sense only if the formulas in the set don't have a symbol s that occurs in one formula as a function symbol of arity n but in another formula as a function symbol of arity m not equal n, or as a relation symbol; and mutatis mutandis for relation symbols. .

By "the language of set theory" we mean the language determined by the signature <0 {'e' '="} {<'e' 2> <'=' 2>}>. (Note: the appearance of '0' there does not imply that '0' is a symbol in the language, but rather that the set of function symbols is empty). This language has no function symbols and only two relation symbols: 'e' and '='. Any other function symbols or relation symbols used for doing set theory are not in the language for set theory but rather they are in a language of set theory extended by definitions. With the method of definitions, any formula that has defined symbols can be reverted mechanically to a certain formula that does not have the defined symbols.

(4)

Df: a model M for a language L with signature <F B t> is a tuple <U V> such that U is a nonempty set and V is a function on FuB that assigns to each f in F a t(f)-place function on U, and to each R in B a t(R)-place relation on U.

So for first order logic with identity: Except for the relation symbol '=', V may map any function symbol to any function on U as long as that function is the arity for the function symbol, and V may map any relation symbol to any relation on U as long as that relation is the arity for the relation symbol. And '=' is always assigned the identity relation on U.

So a model M for the language of set theory is a tuple <U V> such that U is a nonempty set and V('e') is a 2-place relation on U.

Nota bene: A model for the language of set theory is not required to map 'e' to the membership relation on U. If a model M doesn't map 'e' to the membership relation, then M does not adhere to our intuitions of what set theory is about, but M is still a model for the language of set theory.

(5) Df: a sentence P is true in a model M for a language L iff [fill in the recursive definition here that is too detailed for this post].

Df: a set of sentences T has a model iff there is a model M for the language of T such that every member of T is true in M.

Df. a sentence P has a model iff there is a model M for a language for {P} such that P is true in M.

So "ExAy yex" has a model iff there is a model M for a language with 'e' such that "ExAy yex" is true in M.

If a sentence has a model then there is no upper bound to the number of models the sentence has, so, a fortiori, if a sentence has a model then it has at least two models. So "the sentence has a model" implies "the sentence has models". Moreover, I will trivially show two models for "ExAy yex" anyway. I mention this only to ward against a quibble that originally I said 'models' plural.

(5) If a sentence has a model, then the sentence is consistent. So to prove that a sentence is consistent, it suffices to prove that the sentence has a model.

(6) There is a model M such that "ExAy yex" is true in M.

'=' does not occur in "ExAy yex". So a model of "AxEy yex" is:

<{0} {<'e' {<0 0>}>}>

I did not fill in the definition of 'true in the model' previously in this post, because it is too detailed for this post. But here is an intuitive account regarding the above:

The universe is {0}.

The symbol 'e' maps to the relation {<0 0>}.

Nota bene: This is not an interpretation of 'e' that we have in mind for our intuitive meaning of 'e'. But a model does not have to conform to our intuitive meanings of the symbols. For the purpose of modeling the sentence, we can interpret 'e' as standing for any 2-place relation on the domain.

Uninterpreted, "ExAy yex" says that there is an x in whatever is the domain, such that every y in whatever is the domain bears whatever is the relation symbolized by 'e' to x.

With the interpretation above, the domain is {0} and the relation symbolized by 'e' is {<0 0>}.

And every y in the domain (the only y in the domain is 0) bears the relation {<0 0>} to 0. So there is an x (viz. 0) in the domain, such that every y in the domain bears the relation {<0 0>} to x.

Or, including '=' as a symbol, here is a model for the language of set theory that is a model of "AxEy yex":

<{0} {<'e' {<0 0>}> <'=' {<0 0>}>}>

Nota bene: Trivially there are theorems of set theory that are false in this model. So this model is not a model of set theory. My claim has never been that there is a model of "ExAy yex" that is a model of set theory. The model I show is a model for the language of set theory, but it is not a model of set theory. It doesn't need to be a model of set theory.

And to make it 'models' plural, trivially here's another:

<{1} {<'e' {<1 1>}> <'=' {<1 1>}>}>

And another that is not isomorphic with those:

<{0 1} {<'e' {<0 1> <1 1>}> <'=' {<0 0>}>}>

those models are in no way sets. — fishfry

'those models' there refers to models I claimed to exist, I but I had not specified them.

The domains of the models are sets. And the models themselves are sets:

For example, <{0} {<'e' {<0 0>}>}> is an ordered pair, and ordered pairs are sets.

I will try to learn more about it through time with more practicing — Corvus

Its a good bet that, if you're not taking a class, then the best way to learn is from a good textbook.

'Logic: Techniques of Formal Reasoning' by Kalish, Montague, and Mar is the best introduction, in my opinion based on having looked at a lot of logic books.

symbolic logic can be a bit inadequate for arriving at true conclusions, even if the arguments look valid and consistent. — Corvus

You gave examples of arguments that symbolic logic rules as invalid. That's not a problem for symbolic logic; it's only a problem for you if you think symbolic logic does rule those arguments as valid.

All d -> a (true)
All c -> a (true)
Therefore c = d (true)

Above logical arguments look OK in the symbols. But when they are put in with the real objects in the world, it leads to the wrong conclusion.

All dogs are animals. (true)
All cats are animals. (true)
Therefore cats are dogs (false)

What are the actual problems here? — Corvus

The problems are:

(1) Your first example is not correct syntax, and even when corrected, it is irrelevant

Maybe you meant:

Ax(Dx -> Bx)
Ax(Cx -> Bx)
therefore Ax(Cx <-> Dx)

But the logical calculus doesn't permit that inference so your example is irrelevant.

Or more simply you might mean:

D -> B
C -> A
therefore C <-> D

Again, the logical calculus doesn't permit that inference so your example is irrelevant.

(2) Your second argument is not correct syntax, and even when corrected, it is irrelevant.

Maybe you mean:

Ax(Dx -> Nx)
Ax(Cx -> Nx)
therefore Ax(Cx -> Dx)

The logical calculus doesn't permit that inference so your example is irrelevant.

it reads as if when one asserts ~Jc, one has established (Sc v Wc) assuming the argument is valid. — fdrake

Of course, that argument establishes its conclusion only if the premises themselves are established. I don't see a problem.

I take it that you intend (4) as a conclusion from the premises above it.

You end up transitioning to a space of interpretations that excludes juiciness. — fdrake

I don't know what you mean by "space of interpretations that excludes". It would help if you said it in ordinary terminology for logic.

The predicates are 'is sour', 'is sweet' and 'is juicy'. I guess you mean that your intended interpretation has as its domain the set of apples?

Let 'S' stand for 'is sour', 'W' for 'is sweet' and 'J' for 'is juicy'. Let 'c' be a constant.

If the domain is intended to be the set of apples, then we don't need to symbolize 'is an apple'.

So perhaps this captures your argument:

1. Ax(Sx v Wx v Jx) premise
2. Ax(~Jx -> (Sx v Wx)) from 1
3. ~Jc premise
4 Sc v Wc from 2,3

I don't see a problem.

Or, if the intended domain is not specified, and we have 'P' for 'is an apple':

1. Ax(Px -> (Sx v Wx v Jx)) premise
2. Ax(Px -> (~Jx -> (Sx v Wx))) from 1
3. Pc
4. ~Jc premise
5. Sc v Wc from 2,3,4

I don't see a problem.

/

In my predicate argument about the election, let the intended domain be {Carter, Reagan, Anderson}.

That would make some of the background premises unneeded, but logically I don't see a problem.

if it is not Reagan who wins, it will be Jimmy Carter — Banno

C <-> ~(R v A) is a given

~ R -> (C v A) is a given

C <-> ~A is a given

Lets' say:

~R -> C is a given

Then:

(R v A) -> (~R -> A)
(R v A)
therefore (~R -> A) & (~R -> C)

No contradiction.

So the argument above is:

1. (A ∧ B) → C
2. A
3. B → C — Michael

No, it's:

(R v A) -> (~R -> A)
(R v A)
therefore ~R -> A

But the author might argue this:

If modus ponens is valid, then if we believe the premises, then we believe the conclusion (not always in fact - people err - but in principle). And the premises are believed but the conclusion is not. So, still, there's a puzzle.

I essentially agree. My point is that:

(R v A) -> (~R -> A)
R v A
therefore ~R -> A

is an instance of modus ponens, but

(R v A) -> (~R -> A)
R v A
therefore we have reason to believe ~R -> A

is not.

Make explicit 'Republican' and 'Democrat':

'P' stands for 'Republican wins'.
'D' stands for 'Democrat wins'.

Add the background premises:

P <-> (R v A)
D <-> C
R <-> ~(A v C)
A <-> ~(R v C)
C <-> ~(R v A)
R

So, these follow:

P -> (~R -> A)
P

and conclusion:

~R -> A

/

Or spell it out with constants and predicates.

'c' stands for Carter
'r' stands for Reagan
'a' stands for Anderson

'R' stands for 'is a Republican candidate'
'D' stands for 'is a Democratic candidate'
'W' stands for 'wins the election:

Add the background premises:

Ax(Rx <-> (x = r v x = a))
Ax(Dx <-> x = c) [but not needed for the argument]
Rr
Ra
Dc [but not needed for the argument]
Wa <-> ~(Wr v Wc)
Wc <-> ~(Wr v Wa)
Wr <-> ~(Wa v Wc)
Wr [but not needed for the argument]

So, these follow:

1. Ex(Rx & Wx) -> (~Wr -> Wa) from background premises
2. Ex(Rx & Wx) from background premises
3. (Rr & Wr) v (Ra & Wa) from 2 and background premises
4. ((Rr & Wr) v (Ra & Wa)) -> (~Wr -.> Wa)
5. ~Wr -> Wa from 3,4 MP

So even with the finer analysis with constants and predicates, we still arrive at MP captured more easily anyway with just sentence letters.

So my point in response to you is that In a context of classical logic, if an argument is valid then it doesn't become invalid by adding premises of finer analysis (such as predicate logic is finer than propositional logic). This is the monotonic property of classical logic.

Yes, but the puzzle (if there truly is one) maintains with or without resort to talking about domains.

Include the premises:

R
~C
~A

Then

(R v A) -> (~R -> A)
R v A
~A
~C
therefore ~R -> A

Still valid.

Yes, my mistake.

Premise 2 is false — Amalac

At a point before the election, with 'wins' understood as 'will win', then R v A is true.

At a point after the election, with 'wins' understood as 'won', then R v A is true.

Keep track of domains. — fdrake

We may state these atomic propositions purely as sentence letters so there is not a need to involve domains.

Since R is the case, ~R -> A is true.

Bu the puzzle includes an intensional operator "believe'.

'R' for 'Reagan wins'
'A' for 'Anderson wins'
'C' for 'Carter wins'
'R v A' for 'a Republican wins'

(R v A) -> (~R -> A)
R v A
therefore ~R -> A

That's an instance of modus ponens..

What are your thoughts on this?:

https://thephilosophyforum.com/discussion/comment/565970

Maybe there's a paraconsistent modal logic system, and semantics for it that provide that LNC is true in some models but not in other models?

Too hard trying to find where he made the claim about LNC. That's okay, I guess it doesn't bear on this discussion.

one version fo my OP would be to ask if (x) ▢(E!(x)) - if there is some being that exists in all possible worlds. — Banno

I'll rewrite that in text only ('A' for universal quantifier, 'N' for 'necessarily', 'X' for the existence predicate.

Ay N Xy

But what do you mean by "if"? Are you asking whether it's a theorem of some given system? (That system wouldn't be S4 or S5, since those are merely propositional logic systems, or do you mean a quantified version of S4 or S5?) Or are you asking whether there are models in which the formula is true?

Why not this trivial model?

w has domain = {0}
W = {w}
R = {<w w>}
f = {'c' 0}

'E!' is being used as a 1-place predicate symbol.
— TonesInDeepFreeze — Banno

Since that post, I am reading to find more about the existence predicate (that I would call just 'E' and not 'E!' as others have), but I haven't yet caught up to seeing exactly how it works.

Bartricks's claim that LNC is true but contingent — Banno

Would you please link to it?