0.00/0.01 YES 0.00/0.01 0.00/0.01 Problem 1: 0.00/0.01 0.00/0.01 (VAR X Y Z) 0.00/0.01 (STRATEGY CONTEXTSENSITIVE 0.00/0.01 (filter 1 2) 0.00/0.01 (from 1) 0.00/0.01 (head 1) 0.00/0.01 (if 1) 0.00/0.01 (primes) 0.00/0.01 (sieve 1) 0.00/0.01 (tail 1) 0.00/0.01 (0) 0.00/0.01 (cons 1) 0.00/0.01 (divides 1 2) 0.00/0.01 (false) 0.00/0.01 (s 1) 0.00/0.01 (true) 0.00/0.01 ) 0.00/0.01 (RULES 0.00/0.01 filter(s(s(X)),cons(Y,Z)) -> if(divides(s(s(X)),Y),filter(s(s(X)),Z),cons(Y,filter(X,sieve(Y)))) 0.00/0.01 from(X) -> cons(X,from(s(X))) 0.00/0.01 head(cons(X,Y)) -> X 0.00/0.01 if(false,X,Y) -> Y 0.00/0.01 if(true,X,Y) -> X 0.00/0.01 primes -> sieve(from(s(s(0)))) 0.00/0.01 sieve(cons(X,Y)) -> cons(X,filter(X,sieve(Y))) 0.00/0.01 tail(cons(X,Y)) -> Y 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(s(s(X)),cons(Y,Z)) -> if(divides(s(s(X)),Y),filter(s(s(X)),Z),cons(Y,filter(X,sieve(Y)))) 0.00/0.01 from(X) -> cons(X,from(s(X))) 0.00/0.01 head(cons(X,Y)) -> X 0.00/0.01 if(false,X,Y) -> Y 0.00/0.01 if(true,X,Y) -> X 0.00/0.01 primes -> sieve(from(s(s(0)))) 0.00/0.01 sieve(cons(X,Y)) -> cons(X,filter(X,sieve(Y))) 0.00/0.01 tail(cons(X,Y)) -> Y 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 IF(false,X,Y) -> Y 0.00/0.01 IF(true,X,Y) -> X 0.00/0.01 PRIMES -> FROM(s(s(0))) 0.00/0.01 PRIMES -> SIEVE(from(s(s(0)))) 0.00/0.01 TAIL(cons(X,Y)) -> Y 0.00/0.01 -> Rules: 0.00/0.01 filter(s(s(X)),cons(Y,Z)) -> if(divides(s(s(X)),Y),filter(s(s(X)),Z),cons(Y,filter(X,sieve(Y)))) 0.00/0.01 from(X) -> cons(X,from(s(X))) 0.00/0.01 head(cons(X,Y)) -> X 0.00/0.01 if(false,X,Y) -> Y 0.00/0.01 if(true,X,Y) -> X 0.00/0.01 primes -> sieve(from(s(s(0)))) 0.00/0.01 sieve(cons(X,Y)) -> cons(X,filter(X,sieve(Y))) 0.00/0.01 tail(cons(X,Y)) -> Y 0.00/0.01 -> Unhiding Rules: 0.00/0.01 filter(s(s(X)),Z) -> FILTER(s(s(X)),Z) 0.00/0.01 filter(s(s(X)),x3) -> x3 0.00/0.01 filter(X,sieve(Y)) -> FILTER(X,sieve(Y)) 0.00/0.01 filter(X,sieve(Y)) -> SIEVE(Y) 0.00/0.01 filter(X,sieve(x3)) -> x3 0.00/0.01 from(s(X)) -> FROM(s(X)) 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 IF(false,X,Y) -> Y 0.00/0.01 IF(true,X,Y) -> X 0.00/0.01 PRIMES -> FROM(s(s(0))) 0.00/0.01 PRIMES -> SIEVE(from(s(s(0)))) 0.00/0.01 TAIL(cons(X,Y)) -> Y 0.00/0.01 -> Rules: 0.00/0.01 filter(s(s(X)),cons(Y,Z)) -> if(divides(s(s(X)),Y),filter(s(s(X)),Z),cons(Y,filter(X,sieve(Y)))) 0.00/0.01 from(X) -> cons(X,from(s(X))) 0.00/0.01 head(cons(X,Y)) -> X 0.00/0.01 if(false,X,Y) -> Y 0.00/0.01 if(true,X,Y) -> X 0.00/0.01 primes -> sieve(from(s(s(0)))) 0.00/0.01 sieve(cons(X,Y)) -> cons(X,filter(X,sieve(Y))) 0.00/0.01 tail(cons(X,Y)) -> Y 0.00/0.01 -> Unhiding rules: 0.00/0.01 filter(s(s(X)),Z) -> FILTER(s(s(X)),Z) 0.00/0.01 filter(s(s(X)),x3) -> x3 0.00/0.01 filter(X,sieve(Y)) -> FILTER(X,sieve(Y)) 0.00/0.01 filter(X,sieve(Y)) -> SIEVE(Y) 0.00/0.01 filter(X,sieve(x3)) -> x3 0.00/0.01 from(s(X)) -> FROM(s(X)) 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