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Set r = s mod 2.
2. Dovetail steps 3 and 4 until, if ever, step 3 succeeds. If and when step 3 succeeds go to step 5.
 ( Intuitively, if step 3 succeeds in each stage, then  and neither M0 nor M1 Ex*-identifies j p(3). )
3. Search for an extension  of j p(3)[xs] such that  .
4. Let xs, s', denote the least x such that j p(cur)(x) has not been defined before substage s' of stage s.
Go to substage 0.
Begin ( Substage s' )
4.1. Set avail = avail + 1.
Enumerate p(avail) into Wp(2).
For each x < xs, s', set j p(avail) (x) = j p(cur) (x).
4.2. Dovetail steps 4.3 and 4.4 until, if ever, step 4.3 succeeds. If and when step 4.3 succeeds, go to step 4.5.
4.3. Search for an extension  of j p(cur) [xs, s'] such that  .
4.4. For x = xs, s' to  do set j p(avail) (x) = 0.
4.5 Let  be as found in step 4.3.
For each  , set j p(cur) (x) = y if j p(cur) (x) is not already defined.
Set  . (Note the diagonalization.)
Go to substage s' + 1.
End ( Substage s' ) |
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5. Let s be as found in step 3.
For each  , set j p(3)(x) = y if j p(3)(x) is not already defined.
Set  . (Note the diagonalization.)
Go to stage s + 1. |
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Now, consider the following cases. |
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Case 1: Each stage terminates. Let f = j p(3). Clearly, f> is computable and a member of C2 (since f(0) = p(0) and Wp(0) = { p(3) }). Also, because of the success of step 3 and the diagonalization at step 5 in each even (respectively, odd) stage, we have that either M0 (respectively, M1) diverges on f or else the last program output by M0 (respectively, M1) on f commits infinitely many convergent errors. |
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