We consider the following Problem: Strict Trs: { a(b(x)) -> b(a(a(x))) , b(c(x)) -> c(b(b(x))) , c(a(x)) -> a(c(c(x))) , u(a(x)) -> x , v(b(x)) -> x , w(c(x)) -> x , a(u(x)) -> x , b(v(x)) -> x , c(w(x)) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { a(b(x)) -> b(a(a(x))) , b(c(x)) -> c(b(b(x))) , c(a(x)) -> a(c(c(x))) , u(a(x)) -> x , v(b(x)) -> x , w(c(x)) -> x , a(u(x)) -> x , b(v(x)) -> x , c(w(x)) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { w(c(x)) -> x , c(w(x)) -> x} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(a) = {1}, Uargs(b) = {1}, Uargs(c) = {1}, Uargs(u) = {}, Uargs(v) = {}, Uargs(w) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: a(x1) = [1 0] x1 + [1] [1 0] [1] b(x1) = [1 0] x1 + [0] [0 0] [1] c(x1) = [1 0] x1 + [0] [0 1] [0] u(x1) = [1 0] x1 + [0] [0 0] [1] v(x1) = [1 0] x1 + [1] [0 0] [1] w(x1) = [1 0] x1 + [1] [0 1] [1] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { a(b(x)) -> b(a(a(x))) , b(c(x)) -> c(b(b(x))) , c(a(x)) -> a(c(c(x))) , u(a(x)) -> x , v(b(x)) -> x , a(u(x)) -> x , b(v(x)) -> x} Weak Trs: { w(c(x)) -> x , c(w(x)) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { v(b(x)) -> x , b(v(x)) -> x} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(a) = {1}, Uargs(b) = {1}, Uargs(c) = {1}, Uargs(u) = {}, Uargs(v) = {}, Uargs(w) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: a(x1) = [1 0] x1 + [0] [0 0] [0] b(x1) = [1 0] x1 + [0] [0 1] [1] c(x1) = [1 0] x1 + [0] [0 1] [0] u(x1) = [1 0] x1 + [1] [0 0] [1] v(x1) = [1 0] x1 + [1] [0 1] [0] w(x1) = [1 0] x1 + [1] [0 1] [1] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { a(b(x)) -> b(a(a(x))) , b(c(x)) -> c(b(b(x))) , c(a(x)) -> a(c(c(x))) , u(a(x)) -> x , a(u(x)) -> x} Weak Trs: { v(b(x)) -> x , b(v(x)) -> x , w(c(x)) -> x , c(w(x)) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { u(a(x)) -> x , a(u(x)) -> x} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(a) = {1}, Uargs(b) = {1}, Uargs(c) = {1}, Uargs(u) = {}, Uargs(v) = {}, Uargs(w) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: a(x1) = [1 0] x1 + [0] [0 1] [1] b(x1) = [1 0] x1 + [0] [0 1] [0] c(x1) = [1 0] x1 + [0] [0 1] [0] u(x1) = [1 0] x1 + [1] [0 1] [0] v(x1) = [1 0] x1 + [1] [0 1] [0] w(x1) = [1 0] x1 + [1] [0 1] [1] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { a(b(x)) -> b(a(a(x))) , b(c(x)) -> c(b(b(x))) , c(a(x)) -> a(c(c(x)))} Weak Trs: { u(a(x)) -> x , a(u(x)) -> x , v(b(x)) -> x , b(v(x)) -> x , w(c(x)) -> x , c(w(x)) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { a(b(x)) -> b(a(a(x))) , b(c(x)) -> c(b(b(x))) , c(a(x)) -> a(c(c(x)))} Weak Trs: { u(a(x)) -> x , a(u(x)) -> x , v(b(x)) -> x , b(v(x)) -> x , w(c(x)) -> x , c(w(x)) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The problem is match-bounded by 0. The enriched problem is compatible with the following automaton: { a_0(2) -> 1 , b_0(2) -> 1 , c_0(2) -> 1 , u_0(2) -> 1 , v_0(2) -> 1 , w_0(2) -> 1} Hurray, we answered YES(?,O(n^1))