We consider the following Problem: Strict Trs: { minus(minus(x)) -> x , minus(h(x)) -> h(minus(x)) , minus(f(x, y)) -> f(minus(y), minus(x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { minus(minus(x)) -> x , minus(h(x)) -> h(minus(x)) , minus(f(x, y)) -> f(minus(y), minus(x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {minus(minus(x)) -> x} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(minus) = {}, Uargs(h) = {1}, Uargs(f) = {1, 2} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: minus(x1) = [1 0] x1 + [2] [0 1] [1] h(x1) = [1 0] x1 + [0] [0 1] [0] f(x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [1 1] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { minus(h(x)) -> h(minus(x)) , minus(f(x, y)) -> f(minus(y), minus(x))} Weak Trs: {minus(minus(x)) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { minus(h(x)) -> h(minus(x)) , minus(f(x, y)) -> f(minus(y), minus(x))} Weak Trs: {minus(minus(x)) -> x} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The problem is match-bounded by 1. The enriched problem is compatible with the following automaton: { minus_0(2) -> 1 , minus_1(2) -> 3 , h_0(2) -> 2 , h_1(3) -> 1 , h_1(3) -> 3 , f_0(2, 2) -> 2 , f_1(3, 3) -> 1 , f_1(3, 3) -> 3} Hurray, we answered YES(?,O(n^1))