Leibniz' Law and that good old riddle

Question

There exists a Theory of Identity in mathematical logic. I've encountered it for the first time in Principia Mathematica by Alfred North Whitehead and Bertrand Russell (1910).
Quote: "This definition states that $x$ and $y$ are to be called identical when every predicative function satisfied by $x$ is also satisfied by $y$".
Many contemporary philosophers call the principle which expresses this view "Leibniz' Law".
One particularly explicit statement can be found in Introduction to Logic and to the Methodology of Deductive Sciences by Alfred Tarski. In chapter III, On the Theory of Identity, it is read that "Among logical laws which involve the concept of identity, the most fundamental is the following: $x = y$ if, and only if, $x$ and $y$ have every property in common. This law was first stated by Leibniz (although in somewhat different terms)."
Tarski does not provide a reference to the place where, according to him, Leibniz stated that law. Further refinements can be found on the Internet. But, for our purpose, it is sufficient to stick to the original definition, as given with the Theory of Identity by Tarski / Russell and Whitehead: $$ (x = y) :\Longleftrightarrow \left[\;\forall P : P(x) \Longleftrightarrow P(y) \;\right] $$ Where $:\Longleftrightarrow$ means: logically equivalent by definition. Let's try something with that definition. Every property in common, they said. We take that quite literally and have, for example: $$ P(x) :\Longleftrightarrow ( x \, \mbox{is on the left of the} \, "=" \, \mbox{sign} ) $$ With this property in mind, consider the expression: $$ 1 = 1 $$ Then we see that the $1$ on the right in $1 = 1$ is not on the left, hence the property $P(1)$ as defined does not hold for that one. Consequently: $1 \ne 1$. We have run into a paradox.
Oh, you should say, but self-referential properties are of course not allowed. Sure, I am the last one to disagree with you. This highly artificial example stresses an important point, though:

• With Leibniz's Law, almost any but not all properties are in common

The numerosity of these (not self-referential) properties can still be infinite. Let $A$ be a set of properties $P_k$ and let's call $A$ the aspect or scope of the equality (- anybody who knows a better name ? You're quite welcome ! -): $$ A := \left\{ \; P_0(x), P_1(x), P_2(x), \cdots , P_n(x) \; \cdots \right\} $$ Then ($x = y$) shall be pronounced as $x$ is equal to $y$ with respect to $A$, and may optionally be written as: $$ (x \stackrel{A}{=} y) :\Longleftrightarrow \left[\;\forall P_k \in A : P_k(x) \Longleftrightarrow P_k(y) \;\right] $$ It's a matter of routine to prove that common properties of equality (reflexive, symmetric, transitive) are not different with the above modified definition: $$ x \stackrel{A}{=} x \\ (x \stackrel{A}{=} y) \Longrightarrow (y \stackrel{A}{=} x) \\ ((x \stackrel{A}{=} y) \wedge (y \stackrel{A}{=} z)) \Longrightarrow (x \stackrel{A}{=} z) $$ Up to now, we have not been very clear about what sort of properties one should have in mind, when comparing object $x$ with object $y$ in some respect $A$.
Therefore consider the decimal representation of numbers and define the following properties: $$ P_{c,k}(x) \; :\Longleftrightarrow \; \mbox{" cipher at position $k$ in the decimal representation of $x$ is $c$ "} \\ \mbox{where} \quad c \in \{0,1,2,3,4,5,6,7,8,9\} $$ We have the two (in)famous numbers, as announced in the header: $$ 1.000... \quad \mbox{and} \quad 0.999... $$ Indeed, there exist numerous proofs of the following statement ( e.g. Wikipedia ) : $$ 1.000... = 0.999... $$ However, the following statement is easy to prove now as well. So we have run into some sort of a paradox: $$ \neg \left[1.000... \stackrel{A}{=} 0.999... \right] $$ This raises some obvious Questions.
Maybe "common" equality in mathematics is not Leibniz' equality ? But how can that be?
Hasn't equality been rigorously defined with Russel's / Tarski's Theory of Identity ?
Or maybe, is there a difference between identity and equality in mathematics ?
Should $\equiv$ and $\stackrel{A}{=}$ be identified perhaps ? And is the following statement true then: $$ 1.000... = 0.999... \qquad \mbox{but} \qquad 1.000... \not \equiv 0.999... $$

Answer 1

Are you sure that you are not mixing two different concepts ? When you consider decimal representation of real numbers, you are speaking of :

1) rational numbers approximating the real number : in this case $1.000$ and $0.999$ are different rational numbers

2) two names for the same object (the real number $1$) : in this case you are dealing with names as string of symbols of infinite lenght, that are not so "standard".

Added Dec,17.

I think that there are different aspects regarding identity.

1) From a mathematical point of view, equality is not defined; it is assumed. We simply know what equality between numbers, triangles, etc. are.

2) In mathematical logic, first-order logic (see Wikipedia) :

includes the equality symbol as a primitive logical symbol which is always interpreted as the real equality relation between members of the domain of discourse, such that the "two" given members are the same member. This approach also adds certain axioms about equality These equality axioms are:

1.Reflexivity.

2.Substitution for functions.

3.Substitution for formulas.

Many other properties of equality are consequences of the axioms above, for example: symmetry and transitivity.

In second-order logic, the insight of Russell, etc, was that, assuming a formal counterpart of Leibniz's principle, it was possible to derive the aforesaid basic properties of equality. Is this a definition of identity ?

3) And now we have the philosophical issue : what is identity ? According to Quine, second (and higher)-order logic is deeply involved with "ontological committments". So we can hardly say that its definition of equality has answered the philosophical issues related with identity.

Added Dec,17 - part 2

You can read some useful comments in S.C.Kleene, Mathematical Logic (1967) , pag.163 :

In second-order predicate calculus, we can regard $x=y$ as an abbreviation for $\forall P(P(x) \leftrightarrow P(y))$, instead of introducing it as a primitive predicate. Since conceptually the idea of equality underlies our notion of a domain and of predicates over the domain, it seems more elementary and direct to introduce equality as we do than to define it by refernce to all predicates.

You can see also Derek Goldrei, Propositional and Predicate Calculus (Springer, 2005), para.5.4: The equality axioms and non-normal structures, where the limitations of axioms for equality are discussed.

Answer 2

Thanks to some useful comments and an answer (and the downvotes)-: I think now that one of the above definitions is wrong and should be replaced by the following (also mind the additional '.') : $$ P_{c,k}(x) \; :\Longleftrightarrow \; \mbox{" symbol at position $k$ in the decimal representation of $x$ is $c$ "} \\ \mbox{where} \quad c \in \{\mbox{'.'},\mbox{'0'},\mbox{'1'},\mbox{'2'},\mbox{'3'},\mbox{'4'},\mbox{'5'},\mbox{'6'},\mbox{'7'},\mbox{'8'},\mbox{'9'}\} $$ This would mean that the rest of my argument in the question is essentially about character strings representing the numbers and not about the numbers "themselves". Then, of course, it's trivial that: $$ \mbox{'1.000...'} \ne \mbox{'0.999...'} $$

Update. A nice example of the non-triviality of (non)identity is the following. Suppose $x$ and $y$ are pictures, like in well known puzzles for children, when they say "find the differences":

Instead of a single index $(k)$, a double index $(i,j)$ may be preferred here for the predicates $P$ : $$ P_{i,j}(x) \; :\Longleftrightarrow \; \mbox{"pixel in $x$ at position $(i,j)$ is black"} $$ The aspect $A$ of the identity is finite in this case: $$ (x \stackrel{A}{=} y) :\Longleftrightarrow \left[\;\forall P_{i,j} \in A : P_{i,j}(x) \Longleftrightarrow P_{i,j}(y) \;\right] $$ It doesn't matter though how many predicates $P$ are actually involved for the reflexive, symmetric, transitive properties of $\,\stackrel{A}{=}\,$ to hold. So the above pictures are in fact always equal in some respect, even if (quite) some pixels are not the same; just remove the corresponding predicates from the aspect $A$ and you're done.