We consider the following Problem: Strict Trs: { active(U11(tt(), M, N)) -> mark(U12(tt(), M, N)) , active(U12(tt(), M, N)) -> mark(s(plus(N, M))) , active(plus(N, 0())) -> mark(N) , active(plus(N, s(M))) -> mark(U11(tt(), M, N)) , mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3)) , mark(tt()) -> active(tt()) , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3)) , mark(s(X)) -> active(s(mark(X))) , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2))) , mark(0()) -> active(0()) , U11(mark(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, mark(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3) , U11(active(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, active(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, active(X3)) -> U11(X1, X2, X3) , U12(mark(X1), X2, X3) -> U12(X1, X2, X3) , U12(X1, mark(X2), X3) -> U12(X1, X2, X3) , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3) , U12(active(X1), X2, X3) -> U12(X1, X2, X3) , U12(X1, active(X2), X3) -> U12(X1, X2, X3) , U12(X1, X2, active(X3)) -> U12(X1, X2, X3) , s(mark(X)) -> s(X) , s(active(X)) -> s(X) , plus(mark(X1), X2) -> plus(X1, X2) , plus(X1, mark(X2)) -> plus(X1, X2) , plus(active(X1), X2) -> plus(X1, X2) , plus(X1, active(X2)) -> plus(X1, X2)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { active(U11(tt(), M, N)) -> mark(U12(tt(), M, N)) , active(U12(tt(), M, N)) -> mark(s(plus(N, M))) , active(plus(N, 0())) -> mark(N) , active(plus(N, s(M))) -> mark(U11(tt(), M, N)) , mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3)) , mark(tt()) -> active(tt()) , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3)) , mark(s(X)) -> active(s(mark(X))) , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2))) , mark(0()) -> active(0()) , U11(mark(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, mark(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3) , U11(active(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, active(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, active(X3)) -> U11(X1, X2, X3) , U12(mark(X1), X2, X3) -> U12(X1, X2, X3) , U12(X1, mark(X2), X3) -> U12(X1, X2, X3) , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3) , U12(active(X1), X2, X3) -> U12(X1, X2, X3) , U12(X1, active(X2), X3) -> U12(X1, X2, X3) , U12(X1, X2, active(X3)) -> U12(X1, X2, X3) , s(mark(X)) -> s(X) , s(active(X)) -> s(X) , plus(mark(X1), X2) -> plus(X1, X2) , plus(X1, mark(X2)) -> plus(X1, X2) , plus(active(X1), X2) -> plus(X1, X2) , plus(X1, active(X2)) -> plus(X1, X2)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { U12(X1, X2, mark(X3)) -> U12(X1, X2, X3) , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1}, Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: active(x1) = [1 1] x1 + [0] [0 0] [1] U11(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 0] x3 + [0] [0 1] [1 1] [0 1] [0] tt() = [0] [0] mark(x1) = [1 1] x1 + [1] [0 0] [0] U12(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 1] x3 + [0] [0 1] [1 1] [0 0] [0] s(x1) = [1 0] x1 + [0] [0 1] [0] plus(x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 1] [0 1] [0] 0() = [0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { active(U11(tt(), M, N)) -> mark(U12(tt(), M, N)) , active(U12(tt(), M, N)) -> mark(s(plus(N, M))) , active(plus(N, 0())) -> mark(N) , active(plus(N, s(M))) -> mark(U11(tt(), M, N)) , mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3)) , mark(tt()) -> active(tt()) , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3)) , mark(s(X)) -> active(s(mark(X))) , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2))) , mark(0()) -> active(0()) , U11(mark(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, mark(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3) , U11(active(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, active(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, active(X3)) -> U11(X1, X2, X3) , U12(mark(X1), X2, X3) -> U12(X1, X2, X3) , U12(X1, mark(X2), X3) -> U12(X1, X2, X3) , U12(active(X1), X2, X3) -> U12(X1, X2, X3) , U12(X1, active(X2), X3) -> U12(X1, X2, X3) , s(mark(X)) -> s(X) , s(active(X)) -> s(X) , plus(mark(X1), X2) -> plus(X1, X2) , plus(X1, mark(X2)) -> plus(X1, X2) , plus(active(X1), X2) -> plus(X1, X2) , plus(X1, active(X2)) -> plus(X1, X2)} Weak Trs: { U12(X1, X2, mark(X3)) -> U12(X1, X2, X3) , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { U12(X1, mark(X2), X3) -> U12(X1, X2, X3) , U12(X1, active(X2), X3) -> U12(X1, X2, X3) , plus(mark(X1), X2) -> plus(X1, X2) , plus(active(X1), X2) -> plus(X1, X2)} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1}, Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: active(x1) = [1 1] x1 + [1] [0 0] [3] U11(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 0] x3 + [0] [0 1] [1 1] [0 1] [0] tt() = [0] [0] mark(x1) = [1 1] x1 + [1] [0 0] [0] U12(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [0 0] x3 + [0] [0 1] [0 0] [1 1] [0] s(x1) = [1 0] x1 + [0] [0 1] [0] plus(x1, x2) = [1 1] x1 + [1 0] x2 + [0] [0 0] [0 1] [0] 0() = [0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { active(U11(tt(), M, N)) -> mark(U12(tt(), M, N)) , active(U12(tt(), M, N)) -> mark(s(plus(N, M))) , active(plus(N, 0())) -> mark(N) , active(plus(N, s(M))) -> mark(U11(tt(), M, N)) , mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3)) , mark(tt()) -> active(tt()) , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3)) , mark(s(X)) -> active(s(mark(X))) , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2))) , mark(0()) -> active(0()) , U11(mark(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, mark(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3) , U11(active(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, active(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, active(X3)) -> U11(X1, X2, X3) , U12(mark(X1), X2, X3) -> U12(X1, X2, X3) , U12(active(X1), X2, X3) -> U12(X1, X2, X3) , s(mark(X)) -> s(X) , s(active(X)) -> s(X) , plus(X1, mark(X2)) -> plus(X1, X2) , plus(X1, active(X2)) -> plus(X1, X2)} Weak Trs: { U12(X1, mark(X2), X3) -> U12(X1, X2, X3) , U12(X1, active(X2), X3) -> U12(X1, X2, X3) , plus(mark(X1), X2) -> plus(X1, X2) , plus(active(X1), X2) -> plus(X1, X2) , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3) , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { active(U11(tt(), M, N)) -> mark(U12(tt(), M, N)) , active(U12(tt(), M, N)) -> mark(s(plus(N, M))) , active(plus(N, 0())) -> mark(N) , active(plus(N, s(M))) -> mark(U11(tt(), M, N))} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1}, Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: active(x1) = [1 1] x1 + [1] [0 0] [1] U11(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 0] x3 + [0] [0 1] [1 1] [0 1] [0] tt() = [0] [0] mark(x1) = [1 1] x1 + [0] [0 0] [1] U12(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [1 1] x3 + [0] [0 1] [1 1] [0 0] [0] s(x1) = [1 0] x1 + [0] [0 1] [0] plus(x1, x2) = [1 1] x1 + [1 0] x2 + [0] [0 0] [0 1] [0] 0() = [0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3)) , mark(tt()) -> active(tt()) , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3)) , mark(s(X)) -> active(s(mark(X))) , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2))) , mark(0()) -> active(0()) , U11(mark(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, mark(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3) , U11(active(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, active(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, active(X3)) -> U11(X1, X2, X3) , U12(mark(X1), X2, X3) -> U12(X1, X2, X3) , U12(active(X1), X2, X3) -> U12(X1, X2, X3) , s(mark(X)) -> s(X) , s(active(X)) -> s(X) , plus(X1, mark(X2)) -> plus(X1, X2) , plus(X1, active(X2)) -> plus(X1, X2)} Weak Trs: { active(U11(tt(), M, N)) -> mark(U12(tt(), M, N)) , active(U12(tt(), M, N)) -> mark(s(plus(N, M))) , active(plus(N, 0())) -> mark(N) , active(plus(N, s(M))) -> mark(U11(tt(), M, N)) , U12(X1, mark(X2), X3) -> U12(X1, X2, X3) , U12(X1, active(X2), X3) -> U12(X1, X2, X3) , plus(mark(X1), X2) -> plus(X1, X2) , plus(active(X1), X2) -> plus(X1, X2) , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3) , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { U11(mark(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, mark(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3) , U11(active(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, active(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, active(X3)) -> U11(X1, X2, X3) , U12(mark(X1), X2, X3) -> U12(X1, X2, X3) , U12(active(X1), X2, X3) -> U12(X1, X2, X3) , s(mark(X)) -> s(X) , s(active(X)) -> s(X) , plus(X1, mark(X2)) -> plus(X1, X2) , plus(X1, active(X2)) -> plus(X1, X2)} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1}, Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: active(x1) = [1 0] x1 + [2] [1 0] [3] U11(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0] [0 0] [0 0] [0 0] [1] tt() = [2] [0] mark(x1) = [1 0] x1 + [1] [1 0] [1] U12(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1] [0 0] [0 0] [0 0] [1] s(x1) = [1 0] x1 + [0] [0 0] [1] plus(x1, x2) = [1 0] x1 + [1 0] x2 + [2] [1 0] [0 0] [1] 0() = [0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3)) , mark(tt()) -> active(tt()) , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3)) , mark(s(X)) -> active(s(mark(X))) , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2))) , mark(0()) -> active(0())} Weak Trs: { U11(mark(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, mark(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3) , U11(active(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, active(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, active(X3)) -> U11(X1, X2, X3) , U12(mark(X1), X2, X3) -> U12(X1, X2, X3) , U12(active(X1), X2, X3) -> U12(X1, X2, X3) , s(mark(X)) -> s(X) , s(active(X)) -> s(X) , plus(X1, mark(X2)) -> plus(X1, X2) , plus(X1, active(X2)) -> plus(X1, X2) , active(U11(tt(), M, N)) -> mark(U12(tt(), M, N)) , active(U12(tt(), M, N)) -> mark(s(plus(N, M))) , active(plus(N, 0())) -> mark(N) , active(plus(N, s(M))) -> mark(U11(tt(), M, N)) , U12(X1, mark(X2), X3) -> U12(X1, X2, X3) , U12(X1, active(X2), X3) -> U12(X1, X2, X3) , plus(mark(X1), X2) -> plus(X1, X2) , plus(active(X1), X2) -> plus(X1, X2) , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3) , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: { mark(tt()) -> active(tt()) , mark(0()) -> active(0())} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(active) = {1}, Uargs(U11) = {1}, Uargs(mark) = {1}, Uargs(U12) = {1}, Uargs(s) = {1}, Uargs(plus) = {1, 2} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: active(x1) = [1 2] x1 + [0] [0 0] [0] U11(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [2] [0 0] [0 0] [0 0] [1] tt() = [0] [0] mark(x1) = [1 0] x1 + [1] [0 0] [0] U12(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [2] [0 0] [0 0] [0 0] [1] s(x1) = [1 0] x1 + [0] [0 0] [0] plus(x1, x2) = [1 0] x1 + [1 0] x2 + [3] [0 0] [0 0] [0] 0() = [0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3)) , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3)) , mark(s(X)) -> active(s(mark(X))) , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))} Weak Trs: { mark(tt()) -> active(tt()) , mark(0()) -> active(0()) , U11(mark(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, mark(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3) , U11(active(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, active(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, active(X3)) -> U11(X1, X2, X3) , U12(mark(X1), X2, X3) -> U12(X1, X2, X3) , U12(active(X1), X2, X3) -> U12(X1, X2, X3) , s(mark(X)) -> s(X) , s(active(X)) -> s(X) , plus(X1, mark(X2)) -> plus(X1, X2) , plus(X1, active(X2)) -> plus(X1, X2) , active(U11(tt(), M, N)) -> mark(U12(tt(), M, N)) , active(U12(tt(), M, N)) -> mark(s(plus(N, M))) , active(plus(N, 0())) -> mark(N) , active(plus(N, s(M))) -> mark(U11(tt(), M, N)) , U12(X1, mark(X2), X3) -> U12(X1, X2, X3) , U12(X1, active(X2), X3) -> U12(X1, X2, X3) , plus(mark(X1), X2) -> plus(X1, X2) , plus(active(X1), X2) -> plus(X1, X2) , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3) , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { mark(U11(X1, X2, X3)) -> active(U11(mark(X1), X2, X3)) , mark(U12(X1, X2, X3)) -> active(U12(mark(X1), X2, X3)) , mark(s(X)) -> active(s(mark(X))) , mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))} Weak Trs: { mark(tt()) -> active(tt()) , mark(0()) -> active(0()) , U11(mark(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, mark(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, mark(X3)) -> U11(X1, X2, X3) , U11(active(X1), X2, X3) -> U11(X1, X2, X3) , U11(X1, active(X2), X3) -> U11(X1, X2, X3) , U11(X1, X2, active(X3)) -> U11(X1, X2, X3) , U12(mark(X1), X2, X3) -> U12(X1, X2, X3) , U12(active(X1), X2, X3) -> U12(X1, X2, X3) , s(mark(X)) -> s(X) , s(active(X)) -> s(X) , plus(X1, mark(X2)) -> plus(X1, X2) , plus(X1, active(X2)) -> plus(X1, X2) , active(U11(tt(), M, N)) -> mark(U12(tt(), M, N)) , active(U12(tt(), M, N)) -> mark(s(plus(N, M))) , active(plus(N, 0())) -> mark(N) , active(plus(N, s(M))) -> mark(U11(tt(), M, N)) , U12(X1, mark(X2), X3) -> U12(X1, X2, X3) , U12(X1, active(X2), X3) -> U12(X1, X2, X3) , plus(mark(X1), X2) -> plus(X1, X2) , plus(active(X1), X2) -> plus(X1, X2) , U12(X1, X2, mark(X3)) -> U12(X1, X2, X3) , U12(X1, X2, active(X3)) -> U12(X1, X2, X3)} 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: { active_0(2) -> 1 , U11_0(2, 2, 2) -> 1 , tt_0() -> 2 , mark_0(2) -> 1 , U12_0(2, 2, 2) -> 1 , s_0(2) -> 1 , plus_0(2, 2) -> 1 , 0_0() -> 2} Hurray, we answered YES(?,O(n^1))