0.00/0.01 YES 0.00/0.01 0.00/0.01 Problem 1: 0.00/0.01 0.00/0.01 (VAR M N X Y) 0.00/0.01 (STRATEGY CONTEXTSENSITIVE 0.00/0.01 (filter 1 2 3) 0.00/0.01 (nats 1) 0.00/0.01 (sieve 1) 0.00/0.01 (zprimes) 0.00/0.01 (0) 0.00/0.01 (cons 1) 0.00/0.01 (s 1) 0.00/0.01 ) 0.00/0.01 (RULES 0.00/0.01 filter(cons(X,Y),0,M) -> cons(0,filter(Y,M,M)) 0.00/0.01 filter(cons(X,Y),s(N),M) -> cons(X,filter(Y,N,M)) 0.00/0.01 nats(N) -> cons(N,nats(s(N))) 0.00/0.01 sieve(cons(0,Y)) -> cons(0,sieve(Y)) 0.00/0.01 sieve(cons(s(N),Y)) -> cons(s(N),sieve(filter(Y,N,N))) 0.00/0.01 zprimes -> sieve(nats(s(s(0)))) 0.00/0.01 ) 0.00/0.01 0.00/0.01 Problem 1: 0.00/0.01 0.00/0.01 Innermost Equivalent Processor: 0.00/0.01 -> Rules: 0.00/0.01 filter(cons(X,Y),0,M) -> cons(0,filter(Y,M,M)) 0.00/0.01 filter(cons(X,Y),s(N),M) -> cons(X,filter(Y,N,M)) 0.00/0.01 nats(N) -> cons(N,nats(s(N))) 0.00/0.01 sieve(cons(0,Y)) -> cons(0,sieve(Y)) 0.00/0.01 sieve(cons(s(N),Y)) -> cons(s(N),sieve(filter(Y,N,N))) 0.00/0.01 zprimes -> sieve(nats(s(s(0)))) 0.00/0.01 -> The context-sensitive term rewriting system is an orthogonal system. Therefore, innermost cs-termination implies cs-termination. 0.00/0.01 0.00/0.01 0.00/0.01 Problem 1: 0.00/0.01 0.00/0.01 Dependency Pairs Processor: 0.00/0.01 -> Pairs: 0.00/0.01 ZPRIMES -> NATS(s(s(0))) 0.00/0.01 ZPRIMES -> SIEVE(nats(s(s(0)))) 0.00/0.01 -> Rules: 0.00/0.01 filter(cons(X,Y),0,M) -> cons(0,filter(Y,M,M)) 0.00/0.01 filter(cons(X,Y),s(N),M) -> cons(X,filter(Y,N,M)) 0.00/0.01 nats(N) -> cons(N,nats(s(N))) 0.00/0.01 sieve(cons(0,Y)) -> cons(0,sieve(Y)) 0.00/0.01 sieve(cons(s(N),Y)) -> cons(s(N),sieve(filter(Y,N,N))) 0.00/0.01 zprimes -> sieve(nats(s(s(0)))) 0.00/0.01 -> Unhiding Rules: 0.00/0.01 Empty 0.00/0.01 0.00/0.01 Problem 1: 0.00/0.01 0.00/0.01 SCC Processor: 0.00/0.01 -> Pairs: 0.00/0.01 ZPRIMES -> NATS(s(s(0))) 0.00/0.01 ZPRIMES -> SIEVE(nats(s(s(0)))) 0.00/0.01 -> Rules: 0.00/0.01 filter(cons(X,Y),0,M) -> cons(0,filter(Y,M,M)) 0.00/0.01 filter(cons(X,Y),s(N),M) -> cons(X,filter(Y,N,M)) 0.00/0.01 nats(N) -> cons(N,nats(s(N))) 0.00/0.01 sieve(cons(0,Y)) -> cons(0,sieve(Y)) 0.00/0.01 sieve(cons(s(N),Y)) -> cons(s(N),sieve(filter(Y,N,N))) 0.00/0.01 zprimes -> sieve(nats(s(s(0)))) 0.00/0.01 -> Unhiding rules: 0.00/0.01 Empty 0.00/0.01 ->Strongly Connected Components: 0.00/0.01 There is no strongly connected component 0.00/0.01 0.00/0.01 The problem is finite. 0.00/0.01 EOF