Consider the TRS R consisting of the rewrite rules 1: from(X) -> cons(X,n__from(s(X))) 2: 2ndspos(0,Z) -> rnil 3: 2ndspos(s(N),cons(X,Z)) -> 2ndspos(s(N),cons2(X,activate(Z))) 4: 2ndspos(s(N),cons2(X,cons(Y,Z))) -> rcons(posrecip(Y),2ndsneg(N,activate(Z))) 5: 2ndsneg(0,Z) -> rnil 6: 2ndsneg(s(N),cons(X,Z)) -> 2ndsneg(s(N),cons2(X,activate(Z))) 7: 2ndsneg(s(N),cons2(X,cons(Y,Z))) -> rcons(negrecip(Y),2ndspos(N,activate(Z))) 8: pi(X) -> 2ndspos(X,from(0)) 9: plus(0,Y) -> Y 10: plus(s(X),Y) -> s(plus(X,Y)) 11: times(0,Y) -> 0 12: times(s(X),Y) -> plus(Y,times(X,Y)) 13: square(X) -> times(X,X) 14: from(X) -> n__from(X) 15: activate(n__from(X)) -> from(X) 16: activate(X) -> X There are 15 dependency pairs: 17: 2ndspos#(s(N),cons(X,Z)) -> 2ndspos#(s(N),cons2(X,activate(Z))) 18: 2ndspos#(s(N),cons(X,Z)) -> ACTIVATE(Z) 19: 2ndspos#(s(N),cons2(X,cons(Y,Z))) -> 2ndsneg#(N,activate(Z)) 20: 2ndspos#(s(N),cons2(X,cons(Y,Z))) -> ACTIVATE(Z) 21: 2ndsneg#(s(N),cons(X,Z)) -> 2ndsneg#(s(N),cons2(X,activate(Z))) 22: 2ndsneg#(s(N),cons(X,Z)) -> ACTIVATE(Z) 23: 2ndsneg#(s(N),cons2(X,cons(Y,Z))) -> 2ndspos#(N,activate(Z)) 24: 2ndsneg#(s(N),cons2(X,cons(Y,Z))) -> ACTIVATE(Z) 25: PI(X) -> 2ndspos#(X,from(0)) 26: PI(X) -> FROM(0) 27: PLUS(s(X),Y) -> PLUS(X,Y) 28: TIMES(s(X),Y) -> PLUS(Y,times(X,Y)) 29: TIMES(s(X),Y) -> TIMES(X,Y) 30: SQUARE(X) -> TIMES(X,X) 31: ACTIVATE(n__from(X)) -> FROM(X) The approximated dependency graph contains 3 SCCs: {27}, {29} and {17,19,21,23}. - Consider the SCC {27}. There are no usable rules. By taking the polynomial interpretation [s](x) = x + 1 and [PLUS](x,y) = x + y + 1, rule 27 is strictly decreasing. - Consider the SCC {29}. There are no usable rules. By taking the polynomial interpretation [s](x) = x + 1 and [TIMES](x,y) = x + y + 1, rule 29 is strictly decreasing. - Consider the SCC {17,19,21,23}. The usable rules are {1,14-16}. By taking the polynomial interpretation [n__from](x) = x, [2ndsneg#](x,y) = [2ndspos#](x,y) = [activate](x) = [from](x) = [s](x) = x + 1, [cons2](x,y) = x + y + 1 and [cons](x,y) = y, the rules in {1,15,17,21} are weakly decreasing and the rules in {14,16,19,23} are strictly decreasing. Hence the TRS is terminating.