Some of the forms of argument that aren't on the list will probably be complex combinations chosen from the above, however there may also be fresh types.

Arguments are dependent, for usefulness, on being "understood".

Nonetheless there are also many kinds of fallacies and textbooks can list those.

Then there will in practice be the employment of combinations of sound and unsound arguments. The more complicated the "better" in some people's eyes.

Rhetoric is a very interesting field of study altogether. Rhetoric can be good. J H Newman and S J Gould were skilled rhetoricians.

Some forms of argument can be used both soundly and unsoundly. For example, under "argument from authority": "Dr J Smith, biology professor, says the British political system will not last more than 30 years in its present shape." It was alleged this was asking to be shot down on the "grounds" that it wasn't his field. BUT I say, he is just as qualified to express this opinion (as every observant member of this society), indeed I see he is a wide and deep thinker about "political animals"!

:up:

I have got hold of Stanley Jevons and a couple of other decent books. When I get a chance I'll list lots more. There are dozens upon dozens!

I wish I'd been told these things when I was a youngster.

1. Any and all possible logical arguments — TheMadFool

It would be an exercise, but I think even some on your list are reducible to one or more of the others. I even think (don't know, could be wrong) that all can be reduced to negation, "~," and or, "v."

For example: given ~, v, and premises a, b: then (in no particular order):
1) a, b, ~a, ~b, a v b, given
2) (a <=> b) <=> ~ ((a ^ ~b) v (a ^ ~b))
3) a->b <=> ~a v b
4) a ^ b <=> ~(~a v ~b)

And so forth.

I'll add this question: truth tables can be fun, but are in any case tedious and time consuming. Is there a better, faster way to test arguments?

I remember vaguely about how any statement can be rephrased as a NEGATION-OR combination.

I'll add this question: truth tables can be fun, but are in any case tedious and time consuming. Is there a better, faster way to test arguments? — tim wood

What about abbreviated truth tables?

Then there's another method where you assume worlds and check if in any world the argument is invalid. I forgot how it's done.

It seems to me that we either see all argumentation as resting on truth tables for the operators/connectives we've defined, in which case stuff like modus ponens wouldn't be a form we'd specify, it's just a possible compound, or the forms would be theoretically endless.

Is our list of valid argument forms complete? — TheMadFool

There is also an alternative version of propositional calculus that only has one inference rule, i.e. modus ponens, and two operators, AND and OR, which allows for axiomatizing the core rewrite rules of Aristotelian/Boolean two-valued logic.

The basic and derived argument forms then naturally follow from this axiomatization of propositional logic.

Two-valued (Aristotelian/Boolean) logic -- with only the TRUE and FALSE values supported -- is actually a degenerate form of logic that allows for single-value negation results. Not all paradoxical results in two-valued logic also occur in many-valued logic. Therefore, two-valued logic often ends up being considered naive and unusable. For example, the IEEE 1164 (9-valued) and IEEE 1364 (4-valued) international standards disallow the use of naive 2-valued logic in electronic design automation.

Is our list of valid argument forms complete? — TheMadFool

It depends on what you mean by "argument forms." As just pointed out, what you seem to be seeking is an axiomatization of classical logic, which typically involves a few primitives and an inference rule. The "existential graphs" of Charles Sanders Peirce are an innovative diagrammatic alternative.

The "existential graphs" of Charles Sanders Peirce are an innovative diagrammatic alternative. — aletheist

I was looking for extended definitions for the term "consistency" when either extending the permissible logic language or the permissible truth values.

In simple propositional calculus, inconsistency arises when the theory is capable of generating both $\phi(x)$ and $\neg \phi(x)$.

In first-order logic, introduction of the universal quantifiers $\forall$ and $\exists$, allows for a new form of inconsistency, i.e. $\omega$-inconsistency, where the theory generates both $\forall x (\phi(x))$ and $\exists x (\neg \phi(x))$.

Higher-order logic allows, for example, quantification over sets: $\forall x \forall A (x \in A \wedge \phi(x))$. What new mischief is possible here?

So, I was looking for what the term "consistency" means in higher-order, many-valued logic. What new forms of mischief are not allowed? What new forms of inconsistency can be introduced by further increasing the power of the logic language?

The Wikipedia page on consistency only mentions Henkin's theorem in that respect, but I frankly do not understand what exactly this theorem means ... ;-)

Thanks for your valuable answers. I must confess I didn't understand everything the links were about.

Anyway...

To put it in logical terms I'm looking for a proof that shows that the given set of argumentation forms (listed in the OP) is

1) the exact (no less no more) set of valid forms we need to construct any and all arguments

2) any and all arguments can be rephrased/reduced to a combination of these forms

I even think (don't know, could be wrong) that all can be reduced to negation, "~," and or, "v." — tim wood

Do you know of a reason why negation and or are preferred over other logical connective like "and" and "implication"?

Is it that "negation" and "or" require the least number of symbols?

Thank you again for your valuable comments. I was reading another thread of mine concerning axioms and @alcontali mentioned Godel's incompleteness theorems which, to my understanding, proves that there will always be some unprovable truths in any axiomatic system.

Therefore, the no set of valid argument forms is ever going to be enough. There will always be an unprovable truth.

Is there a way to distinguish the provable from the unprovable?

be some unprovable truths in any axiomatic system. — TheMadFool

In some systems, not all. Maybe @alcontali can tell us which, and why.

In some systems, not all. Maybe alcontali can tell us which, and why. — tim wood

An example of a system that is weak enough as such that the incompleteness problem does not occur is the Presburger arithmetic. It only uses "0" and "1" as numbers and only addition "+" as operator.

So, it is only capable of saying things of the following form:

1) 0+0=0
2) 0+1=1 and 1+0=1
3) 1+1=0

Universal quantifiers only run over {0,1}. So, in my impression, they may not even be needed. For example, $\forall x$ can be always be replaced by x=0 or x=1 (This is known as quantifier elimination).

The following remark supports that:

The decidability of Presburger arithmetic can be shown using quantifier elimination, supplemented by reasoning about arithmetical congruence (Enderton 2001, p. 188).

Still, there is also the following remark:

Generally, any number concept leading to multiplication cannot be defined in Presburger arithmetic, since that leads to incompleteness and undecidability.

I am not really sure why that is, though. According to the remark above, axiomatizing the following would cause incompleteness. Not sure why, though:

1) 0*0=0
2) 1*0=0 and 0*1=0
3) 1*1=1

I do not completely understand how adding this simple multiplication scheme would prevent quantifier elimination, but the text seems to claim that it does. It has something to do with "arithmetical congruence". (how!?)

So, my take on the matter is that, if quantifier elimination is not possible, because for example that would lead to infinitely long axioms, then the theory really requires first-order logic, and in that case, it will be incomplete.

For example, normal (=Peano) arithmetic runs over an infinite set n $\in$ { 1, 3, 4, ... }. Quantifier elimination would lead to replacing it by sentences that look like: n=1 or n=2 or n=3 or n=4 ... which is an infinitely long sentence. Hence, the use of first-order logic quantifiers cannot be avoided -- to keep the size of the theory's rules finite -- leading to incompleteness.

"For each natural number the following is true" becomes: it is true for 1. It is true for 2. It is true for 3 ... ad infinitum. "There exists a natural number for which the following is true" becomes: It could be true for 1. It could be true for 2. It could be true for 3 ... ad infinitum.

So, my tentative interpretation is that when a theory gets rephrased in propositional logic (=without universal quantifiers), and it will always consist of infinitely-long construction rules, then the theory will be incomplete.

(but issues with "arithmetical congruence" can apparently also cause incompleteness, but I am not finished trying to figure that out yet ...)

:up:

Thank you.

I was just wondering whether Godel's incompleteness theorems proves that any given set of valid argument forms constructed for our convenience is necessarily inadequate since there'll always be some truths that can't be proved at all.

One thing I'd like to know is whether logic - the entire system - is complete or not. I vaguely remember reading somewhere that logic is a complete system.

From WIki:

The language of Presburger arithmetic contains constants 0 and 1 and a binary function +, interpreted as addition. In this language, the axioms of Presburger arithmetic are the universal closures of the following:

"¬(0 = x + 1)
x + 1 = y + 1 → x = y
x + 0 = x
x + (y + 1) = (x + y) + 1
Let P(x) be a first-order formula in the language of Presburger arithmetic with a free variable x (and possibly other free variables). Then the following formula is an axiom:
(P(0) ∧ ∀x(P(x) → P(x + 1))) → ∀y P(y).
(5) is an axiom schema of induction, representing infinitely many axioms. Since the axioms in the schema in (5) cannot be replaced by any finite number of axioms, Presburger arithmetic is not finitely axiomatizable in first-order logic.

Presburger arithmetic cannot formalize concepts such as divisibility or primality. Generally, any number concept leading to multiplication cannot be defined in Presburger arithmetic, since that leads to incompleteness and undecidability. However, it can formulate individual instances of divisibility; for example, it proves "for all x, there exists y : (y + y = x) ∨ (y + y + 1 = x)". This states that every number is either even or odd."
----

I was just wondering whether Godel's incompleteness theorems proves that any given set of valid argument forms constructed for our convenience is necessarily inadequate since there'll always be some truths that can't be proved at all. — TheMadFool

No. At this point, find and read the proof - Godel's 1931 paper. In English it's very readable. And the idea of truths that cannot be proved at all is not what anyone is concerned with. The proofs in question are meta-mathematical, a fancy way of saying that they're proved indirectly, "outside" the system. For example, Godel's odd sentence, which reads "17genr," roughly and inadequately translated says "17genr is not provable." Metamathematically, then, If true, then true and (but) unprovable. If false, then provable, but then a false statement is provable, and thereby every sentence becomes provable. ("17genr" and "17genr is not provable" being exactly the same sentence.)

One thing I'd like to know is whether logic - the entire system - is complete or not. I vaguely remember reading somewhere that logic is a complete system. — TheMadFool

Well, this comes down to defining your system. Presburger arithmetic is apparently provably both complete and consistent. Peano arithmetic on the other hand, not.

WIki again: "Peano arithmetic, which is Presburger arithmetic augmented with multiplication, is not decidable, as a consequence of the negative answer to the Entscheidungsproblem. By Gödel's incompleteness theorem, Peano arithmetic is incomplete and its consistency is not internally provable."

So in sum, the devils are altogether in the details, and amateurs get in trouble when they start thinking, asking, talking, in terms of details.

One thing I'd like to know is whether logic - the entire system - is complete or not. I vaguely remember reading somewhere that logic is a complete system. — TheMadFool

As already noted, it depends on which system of logic you have in mind, since Gödel's incompleteness theorem only applies to formal systems "within which a certain amount of arithmetic can be carried out" (Stanford). For example, there are well-established proofs that the now-standard systems of first-order propositional and predicate logic are both (deductively) complete and consistent.

For example, there are well-established proofs that the now-standard systems of first-order propositional and predicate logic are both (deductively) complete and consistent. — aletheist

You mean it's possible to create an axiomatic system that is complete and consistent as long as it doesn't involve arithmetic?

Do you know why this is the case?

You mean it's possible to create an axiomatic system that is complete and consistent as long as it doesn't involve arithmetic? — TheMadFool

I mean exactly what I said, quoting the Stanford article--Gödel's incompleteness theorem only applies to formal systems "within which a certain amount of arithmetic can be carried out" (emphasis mine).

Do you know why this is the case? — TheMadFool

Because there are minimum requirements for a formal system to be able to generate the kind of undecidable sentence that Gödel's incompleteness theorem requires. gave an example of a formal system that can do some arithmetic, but not enough for the theorem to apply.

:ok: :up:

Thank you for such a valuable information. I was looking for a long time detailed explanation. Really appreciate your reply!

i) 2 truth values.

ii) 4 binary operators/connectives (conjunction, disjunction, implication, double implication)

iii) 1 unary operator (negation i.e. double negation)

How many permutations are possible?

For binary operators, 2 truth values give us 4 (TT, TF, FT, FF).

Each one of the 4 (above) can be permuted with itself and others: 4 $\times$ 4 = 16 permutations.

There are 16 implication-type argument forms. Not all of them are be valid. In fact only 8 are valid.

Come to equivalence now, there's no limit here but it appears that 10 suffices.

Basically, it's a combinatorics question, more or less. I neither have the wherewithal nor the patience to answer this question.

Thanks a lot for the explanation.