We consider the following Problem: Strict Trs: { c(c(b(c(x)))) -> b(a(0(), c(x))) , c(c(x)) -> b(c(b(c(x)))) , a(0(), x) -> c(c(x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { c(c(b(c(x)))) -> b(a(0(), c(x))) , c(c(x)) -> b(c(b(c(x)))) , a(0(), x) -> c(c(x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {a(0(), x) -> c(c(x))} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(c) = {1}, Uargs(b) = {1}, Uargs(a) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: c(x1) = [1 0] x1 + [0] [0 0] [0] b(x1) = [1 0] x1 + [0] [0 0] [1] a(x1, x2) = [0 2] x1 + [1 0] x2 + [0] [0 0] [0 0] [1] 0() = [0] [2] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { c(c(b(c(x)))) -> b(a(0(), c(x))) , c(c(x)) -> b(c(b(c(x))))} Weak Trs: {a(0(), x) -> c(c(x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {c(c(x)) -> b(c(b(c(x))))} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(c) = {1}, Uargs(b) = {1}, Uargs(a) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: c(x1) = [1 1] x1 + [0] [0 0] [1] b(x1) = [1 0] x1 + [0] [0 0] [0] a(x1, x2) = [0 3] x1 + [1 2] x2 + [0] [0 0] [0 0] [1] 0() = [0] [3] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: {c(c(b(c(x)))) -> b(a(0(), c(x)))} Weak Trs: { c(c(x)) -> b(c(b(c(x)))) , a(0(), x) -> c(c(x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: {c(c(b(c(x)))) -> b(a(0(), c(x)))} Weak Trs: { c(c(x)) -> b(c(b(c(x)))) , a(0(), x) -> c(c(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: { c_0(1) -> 3 , c_0(2) -> 1 , c_0(4) -> 1 , c_0(6) -> 5 , b_0(1) -> 6 , b_0(2) -> 2 , b_0(4) -> 2 , b_0(5) -> 3 , a_0(2, 2) -> 3 , a_0(2, 4) -> 3 , a_0(4, 2) -> 3 , a_0(4, 4) -> 3 , 0_0() -> 4} Hurray, we answered YES(?,O(n^1))