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Proof: Suppose by way of contradiction that M TxtEx*-identifies  . Then by the operator recursion theorem, there exists a recursive, 1-1, increasing p such that Wp(.) may be defined in stages as described below. |
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In the following construction,  denotes the part of Wp(i) enumerated before stage s. (Note that there is no stage 0.) Let  and, for each i > 1, let  . Also let s 1 = (<0, p(0)>, <1, p(1)>). Go to stage 1. |
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Begin ( Stage s )
1. Enumerate  into each of Wp(0), Wp(1), Wp(2s) and Wp(2s + 1).
 Enumerate <2, p(2s)> into both Wp(0) and Wp(2s)..
Enumerate <2, p(2s + 1)> into both Wp(1) and Wp(2s + 1).
Let  be an extension of s s such that  .
Let  be an extension of s s such that 
2. Dovetail between 2a and 2b until, if ever, step 2b succeeds. If and when step 2b succeeds, go to step 3.
2a. 
2b. Search for  ,  such that
 .
3. If and when 2b succeeds, let i and n be as found in 2b and set
Let s s+1 = an extension of  ) such that content( s s+1) = S.
Enumerate S in both Wp(0) and Wp(1).
Go to stage s + 1. |
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Consider the following two cases. |
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Case 1: Infinitely many stages are executed. In this case, let  . But M on  , a text for L, makes infinitely many mind changes—since the only |
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