In Mathias' paper Hilbert, Bourbaki and the scorning of logic (see ), he mentions Hilbert's $\varepsilon$-calculus (and thus Bourbaki's $\tau$-calculus) is not suitable for incomplete systems. But he did not elaborate if there is any real problem (like inconsistency or incapability) with it. Obviously Hilbert hoped to use $\varepsilon$-calculus to prove the completeness of systems like ZFC, and failed as we know. But this does not means $\varepsilon$-calculus itself has any problems. So, is  Mathias' criticism out of his personal flavor of logic and educational reasons, or there is any real problem for $\varepsilon$-calculus in incomplete systems?
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In Mathias' paper Hilbert, Bourbaki and the scorning of logic (see ), he mentions Hilbert's $\varepsilon$-calculus (and thus Bourbaki's $\tau$-calculus) is not suitable for incomplete systems. But he did not elaborate if there is any real problem (like inconsistency or incapability) with it. Obviously Hilbert hoped to use $\varepsilon$-calculus to prove the completeness of systems like ZFC, and failed as we know. But this does not mean $\varepsilon$-calculus itself has any problems. So, is  Mathias' criticism out of his personal flavor of logic and educational reasons, or there is any real problem for $\varepsilon$-calculus in incomplete systems?
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The answer seems to be yes, judging from the following exercise I found in the book Mathematical Logic by H.D. Ebbinghaus, J. Flum, and W. Thomas:     Let $S$ be a finite symbol set and let $\mathfrak{U}$ be a finite $S$-structure. Show that there is an $S$-sentence $\varphi _{\mathfrak{U}}$ the models of which are precisely the $S$-structures isomorphic to $\mathfrak{U}$.   I think I have an idea of how to solve this exercise, but I seem to be unable to materialize it. Thanks.
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Please guide me in this problem. I am confused about whether its asking that having the relation $z&gt;0$ does not satisfy the order axioms.  Any help would be really appreciated.  Thanks!
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Is intutionistic propositional logic decidable?  If so, what is a decision procedure for it, like tableaux for classical propositional logic?  EDIT: In the first revision I mistook "predicate" for "propositional".
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I have to prove that this set of logical operators is not functional complete -  $$ \{\lnot,\leftrightarrow\} $$  i've tried implement this set by $ \{\rightarrow,\lor\} $ which is not functional complete too, but didn't success.. thanks !
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I found this proposition and don't see exactly as to why it is true and even more so, why the converse is false:  Proposition 1. The equivalence between the proposition $z \in D$ and the proposition $(\exists x \in D)x = z$ is provable from the definitory equations of the existential quantifier and of the equality relation. If $D = \{t_{1},t_{2},...t_{n}\}$, the sequent $z = t_{1} \vee z = t_{2} \vee z \vee ... z = t_{n}  \vdash z \in D$ is provable from the definition of the additive disjunction $\vee$.  The converse sequent $z \in D \vdash z = t_{1} \vee z = t_{2} \vee ... z = t_{n}$ is not provable.  The author goes on to say:  "We adopt the intuitionistic interpretation of disjunction. With respect to it, one can characterize a particular class of finite sets"  On the first part of the proposition, well I am not sure what the point is as if we take an element z in D, then we could just call this element x and hence this x = z.  Is there something more to this?  On the second part of Proposition 1 since $z = \text{ some } t \in D$ since t is in the set of axioms and $t = z$, then $t \vdash z$ as z is derivable from t.  For the converse if $z \in D$ then why wouldn't z = some $t \in D$?  Is this because of the Incompleteness theorem?  That perhaps there D as a set of axioms has some consequence which can not be proven by the set of axioms in D?  Or perhaps I am way off here.  Any ideas?  Thanks,  Brian
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I have trouble understanding why it is being said that Godel's incompleteness theorem shows that Hilbert's programme is essentially impossible.  So, if I understand Godel's second incompleteness theorem, it says that sufficiently strong theorems such as Peano Arithmetic cannot prove its consistency within itself.  But isn't Hilbert's programme that we prove consistency using finitary means in metatheory? It seems to me that Godel's theorem implies that we cannot prove the consistency of the theory within the formal theory. But what about proving the consistency of the theory using metatheory (which is outside of the formal theory)?  I would appreciate your help and comments!
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I am curious about why the existence of inaccessible cardinals cannot be proven in ZFC.My intuitive proof is:  Suppose we can prove there is an inaccessible cardinal ϰ,then by definition we must show that for any cardinals b&amp;c which satisfying b&lt;ϰ and c&lt;ϰ,then b^c&lt;ϰ.Then if we consider the set S which is the union of all the cardinals less than ϰ and consider the power set of S--P(S).It follows that |P(S)|>=ϰ.This implies we can proof the existence of inaccessible cardinals under ZFC iff we can disprove it.Therefore we cannot prove the existence using ZFC.  Is this the intuitive idea behind the proof?  I only have learnt a little bit in set theory by self-studying.But I don't know any thing about mathematical logic.So I don't know the thing such as the second incompleteness theorem.This statement was found in my set theory book but without prrof it.So any comment and improvement of my work is welcome!
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What is the difference between Gödel's completeness and incompleteness theorems?
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let D be a deductive system. The system's axiom set, A, includes all propositions that are not tautology, and the only rule of inference is $\frac{\alpha \vee \beta}{\alpha \wedge \beta}$. I need to answer:  Show with induction on the proof that $ \vdash \alpha \Longrightarrow \nvDash \alpha $, i.e if $ \alpha $ can be proven from the empty set, then exists a model $ M $ such that for every $ \varphi \in A $, $ M(\varphi)=T $ and $ M(\alpha)=F$. Is the system strongly sound? Is $X\nvDash \alpha \Longrightarrow X\vdash \alpha$? Prove that the empty set isn't consistent in D    I've already asked about this part of the question here but I still have a problem with it, because every axiom, $a\in A$,  $\vdash a$ and $A \models a$, in contradiction to what I'm supposed to prove here. Moreover because the axiom set includes all what isn't tautology it also includes all that is contradiction, which means that there is no model $ M $ such that for every $ \varphi \in A $, $ M(\varphi)=T $ which means for every proposition $\psi$, $A\Longrightarrow \psi$. This one was fairly easy base on (1) the system isn't sound so it can't be strongly sound. I'm not even sure how to approach this, I don't even have an intuition whether this is correct or not. Assuming this is wrong, I need to find a set $X$ and proposition $\alpha$ such that $X\nvDash \alpha$ but $\alpha$ cannot be proven from $X$, and if it's correct I need to show that for every set, $X$, and every proposition $\alpha$ such that $X\nvDash \alpha$, $\alpha$ can be proven from $X$ On the surface this looks pretty easy, both $(\varphi \wedge \psi)$ and $\neg(\varphi \wedge \psi)$ are axioms, as they aren't tautology, hence $\vdash (\varphi \wedge \psi)$ and $\vdash \neg (\varphi \wedge \psi)$ and the empty set isn't consistent but this seems too easy, like I'm missing something.   EDIT: I might have come up with a solution for (3), if $X\nvDash \alpha$ it means that $\alpha$ isn't a tautology and if it isn't a tautology it is an axiom in D and therefore can be proven from $X$
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I was just thinking about unprovability. I just wanted to know if it is possible to make a concrete boundary between provable problems and unprovable problems in a certain axiomatic system.   We know that there is a statement that is true yet unprovable. Then is it possible that a statement is true and unprovable, but the unprovability is unprovable?  We can extend it to a question of "unprovability of unprovability of unprovability", or any finite number of times, or even $\omega$ times....  I have no clue. Has there been any research on this kind of topic?
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I wanted to find a counter example to show that the completeness of the inner product space is necessary in . Please give an example of a bounded linear functional $T$ on an incomplete inner product space $X$ which do not have any inner product representation i.e. there does not exist any $z$ in $X$ s.t. $T(x)= \langle ,z\rangle$ for all $x$ in $X$.
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Why it is not possible to have such argument?  I found some textbook Introduction to Formal Logic and this is from one of the first excercises. This is marked as impossible, but I do not understand why.
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A similar question was asked in .  A paper by Roberto Di Cosmo and Thomas Dufour () asserts in footnote 2 (p.3) that it has been shown "that there are no nontrivial equations for $⟨\mathbb{N}, \mathrm{Ack}(n, \_, \_)⟩$ if $n &gt; 2.$" References make it clear that $\mathrm{Ack}(n,x,y)=H_{n+1}(x,y)$ in the , so the assertion is as follows:     $\text{There are no nontrivial equations for the binary function }(\_\uparrow^y\_)\text{ if }y&gt;1\\ \text{ (i.e., for hyperoperations above exponentiation)}\tag{*}.$   (For this, the above paper cites Charles F. Martin, "Axiomatic bases for equational theories of natural numbers", Notices of the Am. Math. Soc., 19(7):778, 1972, which I have not found online.)  Q.1: What, precisely, is meant by a "nontrivial equation" here?  (The defining recursion $$x\uparrow^{y+1}(z+1)=x\uparrow^y (x\uparrow^{y+1}z),$$ which certainly holds for $y&gt;1$, is evidently considered trivial.)  Q.2: Is the following a valid proof based on assertion $(*)$? ...  Claim:     If $x,y,z$ are integers $\gt 1$, then the value of $\ x\uparrow^y z\ $ uniquely determines $x,y,z$; i.e.,  $$x\uparrow^y z = x'\uparrow^{y'} z' \quad\implies\quad x=x',\ y=y',\ z=z'.\quad \tag{1}$$  (The purpose of restricting the domain to integers $&gt;1$ is to remove obvious counterexamples that arise from equations like $x\uparrow (a\cdot z) = (x\uparrow a)\uparrow z\ \ $ or $\ \ 2\uparrow^y 2=4\uparrow^{y'} 1\ \ $ or $\ \ 1\uparrow^y z=1\uparrow^{y'} z'.$)  Proof:  Suppose, for the sake of contradiction, that there is a counterexample to $(1)$, in which the LHS of the implication holds but not the RHS. Then the LHS is a nontrivial equation for some hyperoperation(s) beyond exponentiation, which is impossible as per $(*)$. QED
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For example $\{\land, \lnot\}$ apparently forms a functionally complete set as it can form any logical expression.  But I don't really know what this means or how you show it. Do we just show that some set is complete and then in the future if we have a new set, show that we can replicate the known operations with the new ones?  Do you have to find these strategically or is there a systematic way?
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Given the set of standard axioms (I'm not asking for proof of those), do we know for sure that a proof exists for all unproven theorems? For example, I believe the Goldbach Conjecture is not proven even though we "consider" it true.  Phrased another way, have we proven that if a mathematical statement is true, a proof of it exists? That, therefore, anything that is true can be proven, and anything that cannot be proven is not true? Or, is there a counterexample to that statement?  If it hasn't been proven either way, do we have a strong idea one way or the other? Is it generally thought that some theorems can have no proof, or not?
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I thought many results in calculus need axiom choice. For example, I thought one needs AC to prove that a bounded sequence in the real line has a convergent subsequence. Recently I was taught that one only needs mathematical induction to prove it. So here are my questions.  Can most results in calculus be proved without AC? If yes, what are the exceptions, to name a few?  Obviously a theorem which uses Zorn's lemma most likely does need AC. So please exclude obvious ones.  Edit By "without AC", I mean without any form of AC, i.e. countable or not, dependent or not. In other words, within ZF.  Edit One of the motivations of my question is as follows. People often unconciously use AC to prove theorems. And it often turns out that their uses of AC are unnecessary. For example, an infinite subset of a compact metric space has a limit point. In his book "Principles of mathematical analysis", Rudin proves this by choosing a suitable neighborhood of every point of the space. He uses AC here, though he doesn't say so. However, since a compact metric space is separable, you can avoid AC to prove this theorem.  Edit I'll make the above statement clearer. You can even avoid countable AC to prove the above theorem. In other words, you can prove it within ZF.  Edit I'll make my questions clearer and more specific. By calculus, I mean classical analysis in Euclidean spaces. Take, for example, Rudin's "Principles of mathematical analysis". Can all the results in this book be proved within ZF? If not, what are the exceptions?