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