A property about Turing machines can be represented as the language of all Turing machines, encoded as strings, that satisfy that property. The property P is about the language recognized by Turing machines if whenever L(M)=L(N) then P contains (the encoding of) M iff it contains (the encoding of) N. The property is non-trivial if there is at least one Turing machine that has the property, and at least one that hasn't.

Proof: Without limitation of generality we may assume that a Turing machine that recognizes the empty language does not have the property P. For if it does, just take the complement of P. The undecidability of that complement would immediately imply the undecidability of P.

In order to arrive at a contradiction, suppose P is decidable, i.e. there is a halting Turning machine B that recognizes the descriptions of Turing machines that satisfy P. Using B we can construct a Turning machine A that accepts the language {(M,w)| M is the description of a Turing machine that accepts the string w}. As the latter problem is undecidable this will show that B cannot exists and P must be undecidable as well.

Let MP be a Turing machine that satisfies P (as P is non-trivial there must be one). Now A operates as follows:

1. On input (M,w), create a (description of a) Turing machine C(M,w) as follows:
   1. On input x, let the Turing machine M run on the string w until it accepts (so if it doesn't accept C(M,w) will run forever).
   2. Next run MP on x. Accept iff MP does.
   Note that C(M,w) accepts the same language as MP if M accepts w; C(M,w) accepts the empty language if M does not accept w.
   Thus if M accepts w the Turing machine C(M,w) has the property P, and otherwise it doesn't.
2. Feed the description of C(M,w) to B. If B accepts, accept the input (M,w); if B rejects, reject.