By Arto Salomaa

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**Example text**

4. Show that if k is a constant, then the instruction \IF X = k THEN P END" can be simulated by a LOOP(1)-program. C The equivalence problem (for programs of a certain type) is the problem of determining for two given programs if they compute the same function. The equivalence problem for Turing machines is easily seen to be undecidable, since it is at least as di cult as the halting problem. On the other hand, the equivalence problem for nite automata is decidable. The question then is this: With respect to equivalence problems, where exactly is the boundary between undecidability and decidability?

S(x) = x + 1, 2. z n(x1 ; : : : ; xn ) = 0, 3. uni(x1 ; : : : ; xn ) = xi , 4. x1 + x2 , 5. x . k, 0; 6. w(x1 ; x2 ) = x0;1 ; xx2 = > 2 0; 7. x DIV k, 8. x MOD k. where k 2 N is an arbitrary constant. 3. A function is LOOP(1)-computable if and only if it is simple. For the direction from right to left, we observe rst that functions (1){(5) above are all clearly LOOP(1)-computable. 6. Show that the function w in item (6) is LOOP(1)-computable. 7. Show that for every k 2 N , the functions x DIV k and x MOD k are LOOP(1)-computable.

The proof strategy is now the following: We will show that if a resolution proof is \too short" then it must contain an error. Suppose there is a resolution proof P for PHPn that is \too short," where for the length of the proof P we only count the number of complex clauses that occur. Suppose this number is less than cn for some constant c > 1 to be determined later. Then a certain greedy algorithm, which we will give below, with the set of complex clauses in P as input will nd a partial assignment S that satis es all of these complex clauses.