We consider the following Problem: Strict Trs: { fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , from(X) -> cons(X) , add(0(), X) -> X , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s()} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , from(X) -> cons(X) , add(0(), X) -> X , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s()} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s()} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(fst) = {}, Uargs(cons) = {}, Uargs(from) = {}, Uargs(add) = {}, Uargs(len) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: fst(x1, x2) = [0 0] x1 + [0 0] x2 + [1] [0 0] [0 0] [1] 0() = [0] [0] nil() = [0] [0] s() = [0] [0] cons(x1) = [0 0] x1 + [0] [0 0] [0] from(x1) = [0 0] x1 + [0] [0 0] [0] add(x1, x2) = [0 0] x1 + [1 0] x2 + [1] [0 0] [0 0] [1] len(x1) = [0 0] x1 + [1] [0 0] [1] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { from(X) -> cons(X) , add(0(), X) -> X} Weak Trs: { fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s()} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {from(X) -> cons(X)} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(fst) = {}, Uargs(cons) = {}, Uargs(from) = {}, Uargs(add) = {}, Uargs(len) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: fst(x1, x2) = [0 0] x1 + [0 0] x2 + [1] [0 0] [0 0] [1] 0() = [0] [0] nil() = [0] [0] s() = [0] [0] cons(x1) = [0 0] x1 + [0] [0 0] [0] from(x1) = [0 0] x1 + [2] [0 0] [0] add(x1, x2) = [0 0] x1 + [1 0] x2 + [1] [0 0] [0 0] [1] len(x1) = [0 0] x1 + [1] [0 0] [1] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: {add(0(), X) -> X} Weak Trs: { from(X) -> cons(X) , fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s()} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {add(0(), X) -> X} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(fst) = {}, Uargs(cons) = {}, Uargs(from) = {}, Uargs(add) = {}, Uargs(len) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: fst(x1, x2) = [1 0] x1 + [0 0] x2 + [1] [0 1] [0 0] [1] 0() = [0] [0] nil() = [0] [0] s() = [0] [0] cons(x1) = [0 0] x1 + [0] [0 0] [0] from(x1) = [0 0] x1 + [0] [0 0] [0] add(x1, x2) = [0 0] x1 + [1 0] x2 + [1] [0 0] [0 1] [1] len(x1) = [0 0] x1 + [1] [0 0] [1] The strictly oriented rules are moved into the weak component. We consider the following Problem: Weak Trs: { add(0(), X) -> X , from(X) -> cons(X) , fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s()} StartTerms: basic terms Strategy: innermost Certificate: YES(O(1),O(1)) Proof: We consider the following Problem: Weak Trs: { add(0(), X) -> X , from(X) -> cons(X) , fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s()} StartTerms: basic terms Strategy: innermost Certificate: YES(O(1),O(1)) Proof: Empty rules are trivially bounded Hurray, we answered YES(?,O(n^1))