TITLE: Is Pigeonhole Principle the negation of Dedekind-infinite?
QUESTION [16 upvotes]: From Wiki, "The Pigeonhole Principle":

In mathematics, the pigeonhole principle states that if n items are
  put into m pigeonholes with n > m, then at least one pigeonhole must
  contain more than one item.

From "The Mathematical Infinite as a Matter of Method," by Akihiro Kanamori:

In 1872 Dedekind was putting together Was sind und was sollen die
  Zahlen?, and he would be the first to define infinite set, with the
  definition being a set for which there is a one-to-one correspondence (bijection?)
  with a proper subset. This is just the negation of the Pigeonhole
  Principle. Dedekind in effect had inverted a negative aspect of finite
  cardinality into a positive existence definition of the infinite.

How can this be? Dedekind's definition of infinite makes no reference to natural numbers or any kind of ordering:
A set $X$ is Dedekind-infinite iff there exists a proper subset $X'$ of $X$ and bijection $f:X'\rightarrow X$  
See follow-up in my answer below.

REPLY [2 votes]: Back in the 1970's, when I was working on programming verification using Pascal and a mechanization of Hoare's axioms,
I tried to prove the following program (or its equivalent in Pascal with assertions) correct:
function php(n: int, f:maps [0:n] to [1:n]) returns (int, int);
for{i=0 to n-1} for{j=i+1 to n} if (f(i) == f(j) ) return (i, j);
return (0, 0);
Essentially, this is an implementation of the pigeon-hole principle.
The key is that the input is a function (f) that maps
[0 to n] to [1 to n]
so that there are distinct i and j such that
f(i) = f(j); this program finds tham.
The problem was to insert the proper assertions in the program
and use the verification generator to prove that the "return(i, j)"
would always eventually be taken.
I eventually managed to do it,
but the theorems involved were far beyond the capabilities
of the theorem prover available then.
I proved them by hand with a lot of effort.
I know that this isn't a proper answer to the question,
but it is somewhat relevant.
Along the same lines,
I also wrote a version of this function
which cached the function values
so the time would be O(n) instead of
O(n$^2$).
The assertions for this version were more complex
and I was not able to prove its correctness.