MHB If S has the same number of elements as T, then T has the same number as S....

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The discussion centers on proving that if two sets S and T have the same number of elements, then T also has the same number of elements as S. The definition of "same number of elements" is established through a pairing function that ensures each element of S corresponds uniquely to an element of T and vice versa. The complexity arises when formalizing this proof, as intuition about set sizes can be misleading without rigorous argumentation. The participants highlight that while the concept may seem obvious, constructing a formal proof requires careful consideration of the properties of functions and pairings. Ultimately, the proof's non-triviality lies in the need for precise logical reasoning rather than intuitive understanding.
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Hi. The proof that I'm working is in the title. The Defition of "same number as" is given as follows: (1)Two sets S and T are said to have the same nuber of elements if each element of S can be paired with precisely one element ofT in such a way that every element ofT is paired with precisely one element of S.

Notation: If \pi is a pairing of the elements of Swith with those of Tand the element sof S is paired in \pi to the element tof T, we shall write s\overbrace{\leftrightarrow}^{\pi}{t} (I don't know how to put the \pi above the \leftrightarrow without the \overbrace).
I don't really understand the book's proof: Since S has the same number of elements as T, we can select a pairing between the elements of Sand T in accordance with (1).We define a pairing as follows: Ifs is paired with t by the selected pairing \pi, then pairt with s. That is if s\overbrace{\leftrightarrow}^{\pi}{t}, thent is paired with s to form the pairing of the elements T with those of S. If the original pairing satisfied (1),then so will the new pairing. Specifically, since\pihad each elementT paired with a unique element of S, then the second pairing also has this property. Therefore, T has the same number of elements asS.

Please help me to understand why this proposition is not trivial, and also the procedure of the proof.
 
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VoNemo19 said:
(I don't know how to put the \pi above the \leftrightarrow without the \overbrace).
\stackrel{\pi}{\leftrightarrow} or \overset{\pi}{\leftrightarrow} gives $\overset{\pi}{\leftrightarrow}$.

VoNemo19 said:
Please help me to understand why this proposition is not trivial
It is trivial to people because we have a great deal of intuition concerning the number of elements in a set and one-to-one correspondences between sets. Some of these intuitions probably come from visual images. If one has to construct a formal proof, for example, in order to convince a computer, which does not have these intuitions, then the task becomes more challenging.

For example, it is an obvious corollary of the pigeonhole principle that if f : {1, ..., m} -> {1, ..., n} and m > n, then there exist i1 and i2 such that 1 ≤ i1 < i2 ≤ m and f(i1) = f(i2). But if we have a formal proof of this fact, we can do more. We can mechanically extract a program that, given f, would compute such i1 and i2. Writing such program from scratch, though not hard, does not seem so trivial, especially if we are not allowed to use arrays. This implies to me that in considering the original fact obvious, we are cheating and coming up with some arguments that are different from a formal mathematical proof.

This problem is still simple enough, and it means that we are expected to provide a rather detailed proof.

VoNemo19 said:
Two sets S and T are said to have the same nuber of elements if each element of S can be paired with precisely one element ofT
This just means that there exists a function $\pi:S\to T$...

VoNemo19 said:
in such a way that every element ofT is paired with precisely one element of S.
Formally, this means

$\forall y\in T\,\exists! x\in S\,\pi(x)=y~(1)$

Here $\exists!x$ means that there exists a unique x. If, given y, we denote the x whose existence is guaranteed in (1) by $\pi'(y)$, then (1) expresses the following two facts.

$\forall y\in T\, \pi(\pi'(y))=y~(2)$
$\forall y\in T\,\forall x'\in S.\,\pi x'=y\to x'=\pi'(y)~(3)$

Now we need to show that $\pi'$ establishes a pairing that shows that T has the same number of elements as S. We already have that $\pi'$ is a function from T to S, so we need to check

$\forall x\in S\,\exists! y\in T\,\pi'(y)=x~(4)$

Of course, given x, we let $y = \pi(x)$. Now the unique existential quantifier results in the following two facts to prove.

$\forall x\in S\,\pi'(\pi(x))=x~(5)$
$\forall x\in S\,\forall y'\in T.\,\pi'(y')=x\to y'=\pi(x)~(6)$

Hmm, the first attempt to prove (5) would be to use (2), but it has functions in the wrong order. It does not seem so trivial now, does it? Hint: (5) is proved using (3) and (6) is proved using (2).
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
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