We consider the following Problem: Strict Trs: { :(x, x) -> e() , :(x, e()) -> x , i(:(x, y)) -> :(y, x) , :(:(x, y), z) -> :(x, :(z, i(y))) , :(e(), x) -> i(x) , i(i(x)) -> x , i(e()) -> e() , :(x, :(y, i(x))) -> i(y) , :(x, :(y, :(i(x), z))) -> :(i(z), y) , :(i(x), :(y, x)) -> i(y) , :(i(x), :(y, :(x, z))) -> :(i(z), y)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { :(x, x) -> e() , :(x, e()) -> x , i(:(x, y)) -> :(y, x) , :(:(x, y), z) -> :(x, :(z, i(y))) , :(e(), x) -> i(x) , i(i(x)) -> x , i(e()) -> e() , :(x, :(y, i(x))) -> i(y) , :(x, :(y, :(i(x), z))) -> :(i(z), y) , :(i(x), :(y, x)) -> i(y) , :(i(x), :(y, :(x, z))) -> :(i(z), y)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { :(x, x) -> e() , :(x, e()) -> x} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(:) = {1, 2}, Uargs(i) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: :(x1, x2) = [1 0] x1 + [1 0] x2 + [1] [0 1] [0 0] [1] e() = [0] [0] i(x1) = [1 0] x1 + [0] [1 0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { i(:(x, y)) -> :(y, x) , :(:(x, y), z) -> :(x, :(z, i(y))) , :(e(), x) -> i(x) , i(i(x)) -> x , i(e()) -> e() , :(x, :(y, i(x))) -> i(y) , :(x, :(y, :(i(x), z))) -> :(i(z), y) , :(i(x), :(y, x)) -> i(y) , :(i(x), :(y, :(x, z))) -> :(i(z), y)} Weak Trs: { :(x, x) -> e() , :(x, e()) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {i(e()) -> e()} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(:) = {1, 2}, Uargs(i) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: :(x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 1] [0 0] [0] e() = [0] [0] i(x1) = [1 0] x1 + [1] [1 0] [1] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { i(:(x, y)) -> :(y, x) , :(:(x, y), z) -> :(x, :(z, i(y))) , :(e(), x) -> i(x) , i(i(x)) -> x , :(x, :(y, i(x))) -> i(y) , :(x, :(y, :(i(x), z))) -> :(i(z), y) , :(i(x), :(y, x)) -> i(y) , :(i(x), :(y, :(x, z))) -> :(i(z), y)} Weak Trs: { i(e()) -> e() , :(x, x) -> e() , :(x, e()) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { :(e(), x) -> i(x) , :(x, :(y, i(x))) -> i(y) , :(x, :(y, :(i(x), z))) -> :(i(z), y) , :(i(x), :(y, x)) -> i(y) , :(i(x), :(y, :(x, z))) -> :(i(z), y)} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(:) = {1, 2}, Uargs(i) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: :(x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 1] [1 0] [0] e() = [2] [0] i(x1) = [1 0] x1 + [0] [1 0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { i(:(x, y)) -> :(y, x) , :(:(x, y), z) -> :(x, :(z, i(y))) , i(i(x)) -> x} Weak Trs: { :(e(), x) -> i(x) , :(x, :(y, i(x))) -> i(y) , :(x, :(y, :(i(x), z))) -> :(i(z), y) , :(i(x), :(y, x)) -> i(y) , :(i(x), :(y, :(x, z))) -> :(i(z), y) , i(e()) -> e() , :(x, x) -> e() , :(x, e()) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {i(i(x)) -> x} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(:) = {1, 2}, Uargs(i) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: :(x1, x2) = [1 0] x1 + [1 0] x2 + [1] [0 1] [1 1] [3] e() = [0] [0] i(x1) = [1 0] x1 + [1] [0 1] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { i(:(x, y)) -> :(y, x) , :(:(x, y), z) -> :(x, :(z, i(y)))} Weak Trs: { i(i(x)) -> x , :(e(), x) -> i(x) , :(x, :(y, i(x))) -> i(y) , :(x, :(y, :(i(x), z))) -> :(i(z), y) , :(i(x), :(y, x)) -> i(y) , :(i(x), :(y, :(x, z))) -> :(i(z), y) , i(e()) -> e() , :(x, x) -> e() , :(x, e()) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {i(:(x, y)) -> :(y, x)} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(:) = {1, 2}, Uargs(i) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: :(x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 1] [0 1] [0] e() = [2] [0] i(x1) = [1 0] x1 + [1] [0 1] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: {:(:(x, y), z) -> :(x, :(z, i(y)))} Weak Trs: { i(:(x, y)) -> :(y, x) , i(i(x)) -> x , :(e(), x) -> i(x) , :(x, :(y, i(x))) -> i(y) , :(x, :(y, :(i(x), z))) -> :(i(z), y) , :(i(x), :(y, x)) -> i(y) , :(i(x), :(y, :(x, z))) -> :(i(z), y) , i(e()) -> e() , :(x, x) -> e() , :(x, e()) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: {:(:(x, y), z) -> :(x, :(z, i(y)))} Weak Trs: { i(:(x, y)) -> :(y, x) , i(i(x)) -> x , :(e(), x) -> i(x) , :(x, :(y, i(x))) -> i(y) , :(x, :(y, :(i(x), z))) -> :(i(z), y) , :(i(x), :(y, x)) -> i(y) , :(i(x), :(y, :(x, z))) -> :(i(z), y) , i(e()) -> e() , :(x, x) -> e() , :(x, e()) -> 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: { :_0(2, 2) -> 1 , e_0() -> 1 , e_0() -> 2 , i_0(2) -> 1} Hurray, we answered YES(?,O(n^1))