User blog:P進大好きbot/List of common mistakes on formal logic appearing in googology

In order to avoid the repetition of the same explanations, I list several common mistakes on formal logic appearing in googology.

= Why is formal logic important in googology? =

Because it helps us to reject Berry's paradox like "My large number is the natural number definable by \(10^{100}\) or less letters! I win!" Of course, formal logic itself is useful for us to argue in a precise manner.

= What does "definition" mean in mathematics? =

A definition of a term such as a large number in formal logic is simply a formula \(D\) with free variable \(x\) such that \(\exists ! x(D)\) is provable under the axiom on which you are working. In particular, you can never define non-trivial term without axioms.

You may say "My large number is the greatest one definable by \(10^{100}\) or less letters in \(\textrm{FOST}\)! I win!" But it is not a definition in any theory written in \(\textrm{FOST}\). You should declare and specify the meta theory and the precise definition in it. I am afraid that you are confounding formulae in the base theory with those in the meta theory.

Also you may name a large number relative to a model \(M\) of the base set theory by a formula. A formula \(D\) with free variable \(x\) in the base set theory names a natural number relative to \(M\) if \(M \models \exists ! x((x \in \mathbb{N}) \wedge D)\)

= What is the meta theory? =

The meta theory is the theory in which the formal language of the theory on which you are working is defined. I need more precisse explanation.

Let \(L\) be a formal language, and \(A\) be a set of axioms written in \(L\). For example, \((L,A)\) is the pair of a formal language and the axiom of group theory, or the pair of the formal language \(\textrm{FOST}\) and the ZFC axiom \(A\) written in \(\textrm{FOST}\). Imagine that you are working on the theory \(T\) based on \((L,A)\). Then the meta theory \(MT\) is the theory in which \(L\) and \(A\) are terms. Namely, the theory \(T\) is coded in \(MT\).

The meta theory is usually arithmetic or set theory, because it possesses the presentablity of formal languages. In arithmetic, you can define a formal language by using the pair function and Goedel correspondence. In set theory, you can do in a similar way in arithmetic, or you can use the free monoid generated by a specific set.

Of course, \(MT\) might also be a coded theory in another theory, say, the meta meta theory \(MMT\). Otherwise, it is described as a naive finite collection of symbols with purely syntax theoretic comparison of sequences.

I note that \(MT\) is fixed before you consider \(T\). Therefore when you fix \(T\), the maximum of the length of the tower of theories \(T,MT,MMT,\ldots\) is fixed. In particular, when you want to define a large number in \(T\), you are not allowed to arrange the length of the tower of meta theories saying like "I take the meta theory, the meta meta theory, and the meta \(\times \infty\) theory! Using them all, I define a great number! I win!". If you want to do so, then declare the rule which allows you to do so. Then other googologists can also enjoy your rule.

= Is every consistent first order theory incomplete? =

No. You need to read the conditions of the incompleteness theorem carefully. For example, it is well-known that \(\textrm{RCF}\) does not satisfy the condition. I am afraid that you are confounding incompleteness theorem with completeness theorem.

Also, there is a consistent first order theory which is not incomplete. Let \(\Sigma\) denote the set of consistent systems of axioms written in a fixed formal language. Then by Zorn's lemma, there is a maximal element \(A\)in \(\Sigma\) with respect to the inclusion relation. If there is a closed formula \(F\) independent of \(A\), \(A \cup \{F\}\) is consistent. It contradicts the maximality of \(A\). Therefore \(A\) is not incomplete.

= Is it possible to deal with natural numbers without axioms? =

No. First of all, see the definition of a definition in formal logic here. Then you notice that if you could define the notion of natural numbers in an appropriate way (e.g. \(0 \neq 1\)) in first order logic without axioms, then you can define it in the same way in any first order logic such as group theory and category theory. It contradicts the obvious fact that they have a model whose underlying set is a singleton.

= What is a model? =

You may say "I use the truth in the model ! It is stronger than the provability! I win!" It might be true, if you understand the definition of the notion of a model well. But sometimes googologists refer to models in mathematically incorrect ways.

There are several definitions of the notion of a model. \(L\) be a formal language of first order set theory, and \(A\) be an appropriate set of axioms of set theory such as ZF or ZFC. Imagine that you are working on the set theory \(T\) based on \((L,A)\). I denote by \(MT\) the meta theory of it.


 * For a set \(\Sigma\) of formulae in \(T\), which is a term of \(MT\), the formula \(M \models \Sigma\) with free variable \(M\) is defined as the formula in \(T\) given as the satisfaction property presented by using the formal language \(L_M\), which is a term of \(T\), associated to \(M\).
 * For a set \(\Sigma\) of formulae in \(T\), which is a term of \(MT\), the formula \(X \models \Sigma\) with definable class \(X\) is defined as the formula in \(MT\) which is the synonym of \(X \vdash \Sigma\).

= Does a model of the base set theory form a model of coded set theory? =

No. First of all, see the definition of a model here. Let \(T\) denote the base theory on which you are working, and \(MT\) the meta theory of \(T\). For a term \(M\) of \(T\), the correspondence sending a formula \(F\) in \(T\) to the formula \(F^M\) relative to \(M\) is defined in \(MT\), but not in \(T\). Therefore you can never consider the satisfaction property of \(M\) against general formulae in a coded theory in \(T\).

For example, if you consider a formal language \(\textrm{FOST}\) of set theory which is a term on \(T\), then you might think that you can consider \(F^M\) for each formula \(F\) written in \(\textrm{FOST}\). However, you can never define a term \(F^M\) of \(MT\) from a term \(F\) of \(T\). It works only when \(F\) is the code of a formula in \(T\). Therefore a model \(M\) of \(T\) does not form a model of ZFC axiom written in \(\textrm{FOST}\).

You might say "Huh? But \(\textrm{FOST}\) is the language of set theory. So every formula written in \(\textrm{FOST}\) is derived from \(T\)!" However, it is wrong. See the example of a formula in the coded theory which is not derived from a formula in the base theory here.

At least, for each formula \(F\) written in \(\textrm{FOST}\), you can define a subset of the set of maps from the set of free variables occurring in \(F\) to \(M\) in a recursive way. It might work as well as \(F^M\), is not itself really the satisfaction property, which should be a formula in \(T\), because it is just a term of \(T\).

= Is every formula in a coded theory a formula in the base theory? =

No. It is not true, even if both theories are set thoeries with the same axioms. Indeed, if you could do so in general, since there is a model \(M\) of \(\textrm{ZFC} + \neg \textrm{Con}(\textrm{ZFC})\) under the assumption of \(\textrm{Con}(\textrm{ZFC})\) in \(MT\) by incompleteness theorem and completeness theorem, and the sequence of formulae forming the proof of \(\perp\) in \(M\) should be derived from a sequence of formulae in \(T\) and also a sequence of formulae in \(MT\). It contradicts the assumption of \(\textrm{Con}(\textrm{ZFC})\) in \(MT\).

What occurs here? So, consider the following formula. \begin{eqnarray*} \textrm{ZFC} + \textrm{Con}(\textrm{ZFC}) \vdash \left(\exists M(M \models \textrm{ZFC} + \neg \textrm{Con}(\textrm{ZFC})) \right) \end{eqnarray*} I call the theory in which \(M\) lives the base theory. Then it is a provable formula in the meta theory. In order to make the formula clearer, I put a suffix on each occurrence of \(\textrm{ZFC}\). \begin{eqnarray*} \textrm{ZFC}_1 + \textrm{Con}(\textrm{ZFC}_2) \vdash \left(\exists M(M \models \textrm{ZFC}_3 + \neg \textrm{Con}(\textrm{ZFC}_4)) \right) \end{eqnarray*} Then it is easy to see that \(\textrm{ZFC}_1\) is a term of the meta meta theory, \(\textrm{ZFC}_2 = \textrm{ZFC}_3\) is a term of the meta theory, and \(\textrm{ZFC}_4\) is a term of the coded theory. More precisely, the formula \(M \models \textrm{ZFC}_3 + \textrm{Con}(\textrm{ZFC}_4)\) in the base theory is an abbreviation of a formula defined by using the formal language \(L_M\) associated to \(M\).

In particular, \(\textrm{Con}(\textrm{ZFC}_2)\) is not the same as \(\textrm{Con}(\textrm{ZFC}_4))\). Such a problem occurs in the same reason why there is a natural number which is not derived from a meta natural number.

= Is it possible to use a model in the definition of a large number? =

Partially yes, but partially no. You can include a model \(M\) in the definition of a large number, as long as either one of the following holds:
 * You are working on set thoery \(T\) and your formula using \(M\) is actually a definition in \(T\) in the sense of this.
 * You are working on \(M\) itself and your formula actually names a natural number relative to \(M\) in the sense of this.

For example, imagine that you are working on set thoery \(T\). The sentence like "I denote by \(n \in \mathbb{N}\) the cardinality of a finite model \(G\) of group thoery" is not a closed formula in \(T\) defining \(n\) as long as the definition of \(G\) is specified in \(T\).

Similarly, even if \(T\) is ZFC set theory, then you can never use a model \(M\) of ZFC axiom by the same reason as long as you are participating in a large number contest with a rule specifying the use of such an \(M\). The sentence like "I denote by \(n \in \mathbb{N}\) (caution: it is not \(\mathbb{N}^M\)) the minimum of the set of natural numbers \(k\) such that there is a formula \(D\) in \(T\) with \(M \models (\exists ! x(D)) \wedge D[k/x]\)" is not a definition in \(T\) as long as the definition of \(M\) is specified.

Of course, if you are working on a model \(M\) of \(T\) but not on \(T\) itself, then you can use \(M\). It depends on the rule of the googology which you are enjoying. In this case, the sentence like "I denote by \(n \in \mathbb{N}^M\) the minimum of the set of natural numbers greater than any natural number \(k\) satisfying some definition \(D\) in \(T\) with length smaller than or equal to \(10^{100}\)" can be easily interpreted to be a definition, as long as the rule allows you to refer to the meta theory \(MT\).

Indeed, let \(T\) denote the base set theory on which you are working, \(\Sigma\) the set of formulae in \(T\) with length smaller than or equal to \(10^{100}\) with free variable \(x\), and \(D\) the formula with free variable \(y\) given by connecting \(\forall x \in \mathbb{N}, ((\exists ! x(\delta)) \to (\delta \to (x < y))\) for each \(\delta \in \Sigma\) by \(\wedge\). They are terms of \(MT\), and \(n\) can be defined as the minimum of the subset of \(\mathbb{N}\) defined by \(D^M\) by the separation axiom.

Also, if you are working on \(T\) and if you are allowed to use a model \(M\) to define a natural number \(n \in \mathbb{N}\) (not \(\mathbb{N}^M\)), then you can interpret your definition in a similar way.

Be careful that you can never use \(M\) as a model of set theory based on a coded formal language \(\textrm{FOST}\), which is a term of \(T\). See this.

= Is a second order theory stronger than first order theory? =

Partially yes, but partially no. It depends on the meaning of the strength, and on what theories you are considering.


 * Second order logic is a generalisation for first order logic. (strong?)
 * Every second order theory is described in a formal language, which is a term of a first order theory. (weak?)
 * It is impossible to characterise the structure of the standard model of natural numbers by the first order Peano arithmetic because of the existence of the transcendental model, while it is possible to characterise it by a second order interpretation of the arithmetic. (strong?)
 * It is possible to use a natural number \(\infty\) relative to a specific model \(N\) of the first order Peano arithmetic which is greater than any natural numbers relative to \(N\) derived from the meta theory, while it is impossible in the case of the second order interpretation. (weak?)
 * If a first order formula in a second order set theory with axioms in which second order variables appear only to describe a axiom schema is provable, then it is also provable in a \(2\)-sorted first order set theory. (weak?)
 * The consistencies of the second order interpretations of the \(2\)-sorted first order set theories \(\textrm{NBG}\) and \(\textrm{MK}\) is provable under the consistency of the first order set theory \(\textrm{ZFC} + \textrm{Inaccessible}\). (weak?)
 * The consistency of the first order Peano arithmetic is provable under the second order \(\textrm{PRA} + \textrm{Ind}(\epsilon_0)\). (strong?)

Therefore you need to declare what you precisely mean. Also the same problem occurs when you refer to higher order logic.

= Is Rayo's number defined without axioms? =

No. See the argument here.

= Can we avoid using large cardinals by regarding \(I\), \(M\), \(K\), and so on as formal constant term symbols ? =

Partially yes, but partially no. It might be possible for you to "describe" the rules of your OCF without using large cardinals. However, if you prove the well-definedness of it, then you often need to assume the existence of large cardinals as an additional axiom. In order to determine whether you actually need large cardinals or not, you have to complete a proof of the well-definedness. Also see the argument here.

= Does a recursive analogue of a large cardinal play a similar role to the original one? =

Partially yes, but partially no. You need to be very careful of the use of a recursive analogue of a large cardinal. See the guideline of the use of large cardinals in the definition of an ordinal notation here.

= Is a value of a computable function independent of large cardinal axioms? =

Usually yes. If its computation process can be presented by explicit syntax-theoretic operators sending a defining formula of a standard natural number to a defining formula of a standard natural number, the output of the computable function with standard input is completely independent of large cardinal axioms, as long as they are consistent. It is simply because of the provability of the behaviour of each computation step. In particular, the same holds for the termination starting from each standard natural number.

On the other hand, it does not imply that a proof of the termination under large cardinal axioms can be reduced to a proof without them. The provability of "the termination of the computation starting from \(n\)" for any standard natural number does not imply the provability of "the termination of the computation starting from \(n\) for any \(n \in \mathbb{N})". If it is unprovable, then it is not a tautology by the incompleteness theorem, i.e. there is a model of \(\textrm{ZFC}\) on which the termination is actually false by the incompleteness theorem.

In this case, the growth rate of the computable function is not well-defined, because it is given by the parial order defined by the eventual domination. You can just compare each value. Of course, you can replace the definition of the growth rate so that it always makes sense whenever you hope so, but then you should explicitly declare the alternative definition. A candidate of an alternative definition of the eventual dominaton \(f < g\) is the meta theoretic statement that there is a defining formula of a standard natural number \(N\) such that for any defining formula of a standard natural number \(n\) with \(N < n\), the inequality \(f(m) < g(m)\) is provable.

If you do not approapriately replace the definition of the growth rate, your analysis is incorrect because you are using the ill-defined notion or implicit axioms stronger than \(\textrm{ZFC}\). Everyone might agree that any analysis heavily based on the strongest axiom \(0 = 1\) is incorrect.

Maybe it is better to give an example of a computable function whose computation process can be presented by explicit syntax-theoretic operators sending a defining formula of a standard natural number to a defining formula of a standard natural number. Remember that every constant function is computable. Then you can understand that the function \begin{eqnarray*} \textrm{CH}(n) := \left\{ \begin{array}{ll} 0 & (\textrm{if } \forall m \in \mathbb{N}, \aleph_m \neq 2^{\aleph_0}) \\ m & (\textrm{if } \aleph_m = 2^{\aleph_0}) \end{array}\right. \end{eqnarray*} is actually a computable function, whose value is not syntax-theoretically described by a defining formula of a standard natural number. If you say "But none can compute it!", then you need to look up the precise definition of the notion of a Turing machine and a computable function. The "computabilty" in the sense in your natural language has nothing to do with the computability of a function. This is a trivial example, but appears when we consider the diagonalisation of computable functions through formal languages of set theory. Please do not say that it is easy to avoid, before reading my effort to define a computale large number using the PTO of \(\textrm{ZFC}\).

I guess that when googologists mention to a computable function, they only consider the case where the correspoding natural number through the universal Turing machine is also described by a model-independent defining formula. Then the computation process is described in purely syntax-theoretic way.

For example, the natural number corresponding to \(\textrm{CH}\) is not described by a defining formula of a standard natural number. I also explicitly use similar syntax-theoretic restriction in the construction of a computale large number using the PTO of \(\textrm{ZFC}\) in order to avoid uncomputability derived from the undecidability of the order between natural numbers defined by computable functions. It is not so easy.

Summary:
 * 1) The value of a computable function is not necessarily independent of large cardinal axioms.
 * 2) The value of a computable function is independent of large cardinals when its computable process is described by syntax-theoretic operators sending a defining formula of a standard natural number to a defining formula of a standard natural number.
 * 3) The termination of a computable function in the case 2 starting from a fixed standard natural number is independent of large cardinal axioms.
 * 4) The provability of the termination of a computable function in the case 2 (starting from any natural number) might heavily depend on large cardinal axioms. If it is not provable without large cardinal axioms, then it is false on a model.
 * 5) The usual definition of the growth rate assumes the totality. Therefore analysis based on a computable function in the case 4 is incorrect as long as you explicitly replace the definition of the growth rate.

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