YES(O(1),O(n^2)) 955.63/297.04 YES(O(1),O(n^2)) 955.63/297.04 955.63/297.04 We are left with following problem, upon which TcT provides the 955.63/297.04 certificate YES(O(1),O(n^2)). 955.63/297.04 955.63/297.04 Strict Trs: 955.63/297.04 { active(f(x)) -> mark(f(f(x))) 955.63/297.04 , f(active(x)) -> active(f(x)) 955.63/297.04 , f(mark(x)) -> mark(f(x)) 955.63/297.04 , f(no(x)) -> no(f(x)) 955.63/297.04 , chk(no(f(x))) -> 955.63/297.04 f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) 955.63/297.04 , chk(no(c())) -> active(c()) 955.63/297.04 , mat(f(x), f(y())) -> f(mat(x, y())) 955.63/297.04 , mat(f(x), c()) -> no(c()) 955.63/297.04 , tp(mark(x)) -> 955.63/297.04 tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) } 955.63/297.04 Obligation: 955.63/297.04 derivational complexity 955.63/297.04 Answer: 955.63/297.04 YES(O(1),O(n^2)) 955.63/297.04 955.63/297.04 The weightgap principle applies (using the following nonconstant 955.63/297.04 growth matrix-interpretation) 955.63/297.04 955.63/297.04 TcT has computed the following triangular matrix interpretation. 955.63/297.04 Note that the diagonal of the component-wise maxima of 955.63/297.04 interpretation-entries contains no more than 1 non-zero entries. 955.63/297.04 955.63/297.04 [active](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 [f](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 [mark](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 [chk](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 [no](x1) = [1] x1 + [1] 955.63/297.04 955.63/297.04 [mat](x1, x2) = [1] x1 + [1] x2 + [0] 955.63/297.04 955.63/297.04 [X] = [0] 955.63/297.04 955.63/297.04 [y] = [0] 955.63/297.04 955.63/297.04 [c] = [0] 955.63/297.04 955.63/297.04 [tp](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 The order satisfies the following ordering constraints: 955.63/297.04 955.63/297.04 [active(f(x))] = [1] x + [0] 955.63/297.04 >= [1] x + [0] 955.63/297.04 = [mark(f(f(x)))] 955.63/297.04 955.63/297.04 [f(active(x))] = [1] x + [0] 955.63/297.04 >= [1] x + [0] 955.63/297.04 = [active(f(x))] 955.63/297.04 955.63/297.04 [f(mark(x))] = [1] x + [0] 955.63/297.04 >= [1] x + [0] 955.63/297.04 = [mark(f(x))] 955.63/297.04 955.63/297.04 [f(no(x))] = [1] x + [1] 955.63/297.04 >= [1] x + [1] 955.63/297.04 = [no(f(x))] 955.63/297.04 955.63/297.04 [chk(no(f(x)))] = [1] x + [1] 955.63/297.04 > [1] x + [0] 955.63/297.04 = [f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.04 955.63/297.04 [chk(no(c()))] = [1] 955.63/297.04 > [0] 955.63/297.04 = [active(c())] 955.63/297.04 955.63/297.04 [mat(f(x), f(y()))] = [1] x + [0] 955.63/297.04 >= [1] x + [0] 955.63/297.04 = [f(mat(x, y()))] 955.63/297.04 955.63/297.04 [mat(f(x), c())] = [1] x + [0] 955.63/297.04 ? [1] 955.63/297.04 = [no(c())] 955.63/297.04 955.63/297.04 [tp(mark(x))] = [1] x + [0] 955.63/297.04 >= [1] x + [0] 955.63/297.04 = [tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.04 955.63/297.04 955.63/297.04 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 955.63/297.04 955.63/297.04 We are left with following problem, upon which TcT provides the 955.63/297.04 certificate YES(O(1),O(n^2)). 955.63/297.04 955.63/297.04 Strict Trs: 955.63/297.04 { active(f(x)) -> mark(f(f(x))) 955.63/297.04 , f(active(x)) -> active(f(x)) 955.63/297.04 , f(mark(x)) -> mark(f(x)) 955.63/297.04 , f(no(x)) -> no(f(x)) 955.63/297.04 , mat(f(x), f(y())) -> f(mat(x, y())) 955.63/297.04 , mat(f(x), c()) -> no(c()) 955.63/297.04 , tp(mark(x)) -> 955.63/297.04 tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) } 955.63/297.04 Weak Trs: 955.63/297.04 { chk(no(f(x))) -> 955.63/297.04 f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) 955.63/297.04 , chk(no(c())) -> active(c()) } 955.63/297.04 Obligation: 955.63/297.04 derivational complexity 955.63/297.04 Answer: 955.63/297.04 YES(O(1),O(n^2)) 955.63/297.04 955.63/297.04 The weightgap principle applies (using the following nonconstant 955.63/297.04 growth matrix-interpretation) 955.63/297.04 955.63/297.04 TcT has computed the following triangular matrix interpretation. 955.63/297.04 Note that the diagonal of the component-wise maxima of 955.63/297.04 interpretation-entries contains no more than 1 non-zero entries. 955.63/297.04 955.63/297.04 [active](x1) = [1] x1 + [1] 955.63/297.04 955.63/297.04 [f](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 [mark](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 [chk](x1) = [1] x1 + [1] 955.63/297.04 955.63/297.04 [no](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 [mat](x1, x2) = [1] x1 + [1] x2 + [0] 955.63/297.04 955.63/297.04 [X] = [0] 955.63/297.04 955.63/297.04 [y] = [0] 955.63/297.04 955.63/297.04 [c] = [0] 955.63/297.04 955.63/297.04 [tp](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 The order satisfies the following ordering constraints: 955.63/297.04 955.63/297.04 [active(f(x))] = [1] x + [1] 955.63/297.04 > [1] x + [0] 955.63/297.04 = [mark(f(f(x)))] 955.63/297.04 955.63/297.04 [f(active(x))] = [1] x + [1] 955.63/297.04 >= [1] x + [1] 955.63/297.04 = [active(f(x))] 955.63/297.04 955.63/297.04 [f(mark(x))] = [1] x + [0] 955.63/297.04 >= [1] x + [0] 955.63/297.04 = [mark(f(x))] 955.63/297.04 955.63/297.04 [f(no(x))] = [1] x + [0] 955.63/297.04 >= [1] x + [0] 955.63/297.04 = [no(f(x))] 955.63/297.04 955.63/297.04 [chk(no(f(x)))] = [1] x + [1] 955.63/297.04 >= [1] x + [1] 955.63/297.04 = [f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.04 955.63/297.04 [chk(no(c()))] = [1] 955.63/297.04 >= [1] 955.63/297.04 = [active(c())] 955.63/297.04 955.63/297.04 [mat(f(x), f(y()))] = [1] x + [0] 955.63/297.04 >= [1] x + [0] 955.63/297.04 = [f(mat(x, y()))] 955.63/297.04 955.63/297.04 [mat(f(x), c())] = [1] x + [0] 955.63/297.04 >= [0] 955.63/297.04 = [no(c())] 955.63/297.04 955.63/297.04 [tp(mark(x))] = [1] x + [0] 955.63/297.04 ? [1] x + [1] 955.63/297.04 = [tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.04 955.63/297.04 955.63/297.04 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 955.63/297.04 955.63/297.04 We are left with following problem, upon which TcT provides the 955.63/297.04 certificate YES(O(1),O(n^2)). 955.63/297.04 955.63/297.04 Strict Trs: 955.63/297.04 { f(active(x)) -> active(f(x)) 955.63/297.04 , f(mark(x)) -> mark(f(x)) 955.63/297.04 , f(no(x)) -> no(f(x)) 955.63/297.04 , mat(f(x), f(y())) -> f(mat(x, y())) 955.63/297.04 , mat(f(x), c()) -> no(c()) 955.63/297.04 , tp(mark(x)) -> 955.63/297.04 tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) } 955.63/297.04 Weak Trs: 955.63/297.04 { active(f(x)) -> mark(f(f(x))) 955.63/297.04 , chk(no(f(x))) -> 955.63/297.04 f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) 955.63/297.04 , chk(no(c())) -> active(c()) } 955.63/297.04 Obligation: 955.63/297.04 derivational complexity 955.63/297.04 Answer: 955.63/297.04 YES(O(1),O(n^2)) 955.63/297.04 955.63/297.04 The weightgap principle applies (using the following nonconstant 955.63/297.04 growth matrix-interpretation) 955.63/297.04 955.63/297.04 TcT has computed the following triangular matrix interpretation. 955.63/297.04 Note that the diagonal of the component-wise maxima of 955.63/297.04 interpretation-entries contains no more than 1 non-zero entries. 955.63/297.04 955.63/297.04 [active](x1) = [1] x1 + [1] 955.63/297.04 955.63/297.04 [f](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 [mark](x1) = [1] x1 + [1] 955.63/297.04 955.63/297.04 [chk](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 [no](x1) = [1] x1 + [1] 955.63/297.04 955.63/297.04 [mat](x1, x2) = [1] x1 + [1] x2 + [0] 955.63/297.04 955.63/297.04 [X] = [0] 955.63/297.04 955.63/297.04 [y] = [0] 955.63/297.04 955.63/297.04 [c] = [0] 955.63/297.04 955.63/297.04 [tp](x1) = [1] x1 + [0] 955.63/297.04 955.63/297.04 The order satisfies the following ordering constraints: 955.63/297.04 955.63/297.04 [active(f(x))] = [1] x + [1] 955.63/297.04 >= [1] x + [1] 955.63/297.04 = [mark(f(f(x)))] 955.63/297.04 955.63/297.04 [f(active(x))] = [1] x + [1] 955.63/297.04 >= [1] x + [1] 955.63/297.04 = [active(f(x))] 955.63/297.04 955.63/297.04 [f(mark(x))] = [1] x + [1] 955.63/297.04 >= [1] x + [1] 955.63/297.04 = [mark(f(x))] 955.63/297.04 955.63/297.04 [f(no(x))] = [1] x + [1] 955.63/297.04 >= [1] x + [1] 955.63/297.04 = [no(f(x))] 955.63/297.04 955.63/297.04 [chk(no(f(x)))] = [1] x + [1] 955.63/297.04 > [1] x + [0] 955.63/297.04 = [f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.04 955.63/297.04 [chk(no(c()))] = [1] 955.63/297.04 >= [1] 955.63/297.04 = [active(c())] 955.63/297.04 955.63/297.04 [mat(f(x), f(y()))] = [1] x + [0] 955.63/297.04 >= [1] x + [0] 955.63/297.04 = [f(mat(x, y()))] 955.63/297.04 955.63/297.04 [mat(f(x), c())] = [1] x + [0] 955.63/297.04 ? [1] 955.63/297.04 = [no(c())] 955.63/297.04 955.63/297.04 [tp(mark(x))] = [1] x + [1] 955.63/297.04 > [1] x + [0] 955.63/297.04 = [tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.04 955.63/297.04 955.63/297.04 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 955.63/297.04 955.63/297.04 We are left with following problem, upon which TcT provides the 955.63/297.04 certificate YES(O(1),O(n^2)). 955.63/297.04 955.63/297.04 Strict Trs: 955.63/297.04 { f(active(x)) -> active(f(x)) 955.63/297.04 , f(mark(x)) -> mark(f(x)) 955.63/297.04 , f(no(x)) -> no(f(x)) 955.63/297.04 , mat(f(x), f(y())) -> f(mat(x, y())) 955.63/297.04 , mat(f(x), c()) -> no(c()) } 955.63/297.04 Weak Trs: 955.63/297.04 { active(f(x)) -> mark(f(f(x))) 955.63/297.04 , chk(no(f(x))) -> 955.63/297.04 f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) 955.63/297.04 , chk(no(c())) -> active(c()) 955.63/297.04 , tp(mark(x)) -> 955.63/297.04 tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) } 955.63/297.04 Obligation: 955.63/297.04 derivational complexity 955.63/297.04 Answer: 955.63/297.04 YES(O(1),O(n^2)) 955.63/297.04 955.63/297.04 We use the processor 'matrix interpretation of dimension 2' to 955.63/297.04 orient following rules strictly. 955.63/297.04 955.63/297.04 Trs: { mat(f(x), f(y())) -> f(mat(x, y())) } 955.63/297.04 955.63/297.04 The induced complexity on above rules (modulo remaining rules) is 955.63/297.04 YES(?,O(n^1)) . These rules are moved into the corresponding weak 955.63/297.04 component(s). 955.63/297.04 955.63/297.04 Sub-proof: 955.63/297.04 ---------- 955.63/297.04 TcT has computed the following triangular matrix interpretation. 955.63/297.04 Note that the diagonal of the component-wise maxima of 955.63/297.04 interpretation-entries contains no more than 1 non-zero entries. 955.63/297.04 955.63/297.04 [active](x1) = [1 2] x1 + [0] 955.63/297.04 [0 0] [0] 955.63/297.04 955.63/297.04 [f](x1) = [1 1] x1 + [0] 955.63/297.04 [0 0] [0] 955.63/297.04 955.63/297.04 [mark](x1) = [1 1] x1 + [0] 955.63/297.04 [0 0] [0] 955.63/297.04 955.63/297.04 [chk](x1) = [1 0] x1 + [0] 955.63/297.04 [0 0] [0] 955.63/297.04 955.63/297.04 [no](x1) = [1 1] x1 + [0] 955.63/297.04 [0 0] [0] 955.63/297.04 955.63/297.04 [mat](x1, x2) = [1 0] x1 + [1 0] x2 + [0] 955.63/297.04 [0 0] [0 0] [0] 955.63/297.04 955.63/297.04 [X] = [0] 955.63/297.04 [0] 955.63/297.04 955.63/297.04 [y] = [2] 955.63/297.04 [2] 955.63/297.04 955.63/297.04 [c] = [0] 955.63/297.04 [0] 955.63/297.04 955.63/297.04 [tp](x1) = [1 0] x1 + [0] 955.63/297.04 [0 0] [2] 955.63/297.04 955.63/297.04 The order satisfies the following ordering constraints: 955.63/297.04 955.63/297.04 [active(f(x))] = [1 1] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 >= [1 1] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 = [mark(f(f(x)))] 955.63/297.05 955.63/297.05 [f(active(x))] = [1 2] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 >= [1 1] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 = [active(f(x))] 955.63/297.05 955.63/297.05 [f(mark(x))] = [1 1] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 >= [1 1] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 = [mark(f(x))] 955.63/297.05 955.63/297.05 [f(no(x))] = [1 1] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 >= [1 1] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 = [no(f(x))] 955.63/297.05 955.63/297.05 [chk(no(f(x)))] = [1 1] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 >= [1 0] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 = [f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 [chk(no(c()))] = [0] 955.63/297.05 [0] 955.63/297.05 >= [0] 955.63/297.05 [0] 955.63/297.05 = [active(c())] 955.63/297.05 955.63/297.05 [mat(f(x), f(y()))] = [1 1] x + [4] 955.63/297.05 [0 0] [0] 955.63/297.05 > [1 0] x + [2] 955.63/297.05 [0 0] [0] 955.63/297.05 = [f(mat(x, y()))] 955.63/297.05 955.63/297.05 [mat(f(x), c())] = [1 1] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 >= [0] 955.63/297.05 [0] 955.63/297.05 = [no(c())] 955.63/297.05 955.63/297.05 [tp(mark(x))] = [1 1] x + [0] 955.63/297.05 [0 0] [2] 955.63/297.05 >= [1 0] x + [0] 955.63/297.05 [0 0] [2] 955.63/297.05 = [tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 955.63/297.05 We return to the main proof. 955.63/297.05 955.63/297.05 We are left with following problem, upon which TcT provides the 955.63/297.05 certificate YES(O(1),O(n^2)). 955.63/297.05 955.63/297.05 Strict Trs: 955.63/297.05 { f(active(x)) -> active(f(x)) 955.63/297.05 , f(mark(x)) -> mark(f(x)) 955.63/297.05 , f(no(x)) -> no(f(x)) 955.63/297.05 , mat(f(x), c()) -> no(c()) } 955.63/297.05 Weak Trs: 955.63/297.05 { active(f(x)) -> mark(f(f(x))) 955.63/297.05 , chk(no(f(x))) -> 955.63/297.05 f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) 955.63/297.05 , chk(no(c())) -> active(c()) 955.63/297.05 , mat(f(x), f(y())) -> f(mat(x, y())) 955.63/297.05 , tp(mark(x)) -> 955.63/297.05 tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) } 955.63/297.05 Obligation: 955.63/297.05 derivational complexity 955.63/297.05 Answer: 955.63/297.05 YES(O(1),O(n^2)) 955.63/297.05 955.63/297.05 We use the processor 'matrix interpretation of dimension 3' to 955.63/297.05 orient following rules strictly. 955.63/297.05 955.63/297.05 Trs: { mat(f(x), c()) -> no(c()) } 955.63/297.05 955.63/297.05 The induced complexity on above rules (modulo remaining rules) is 955.63/297.05 YES(?,O(n^1)) . These rules are moved into the corresponding weak 955.63/297.05 component(s). 955.63/297.05 955.63/297.05 Sub-proof: 955.63/297.05 ---------- 955.63/297.05 TcT has computed the following triangular matrix interpretation. 955.63/297.05 Note that the diagonal of the component-wise maxima of 955.63/297.05 interpretation-entries contains no more than 1 non-zero entries. 955.63/297.05 955.63/297.05 [1 1 1] [0] 955.63/297.05 [active](x1) = [0 0 0] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 1 0] [0] 955.63/297.05 [f](x1) = [0 0 1] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 1 2] [0] 955.63/297.05 [mark](x1) = [0 0 2] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 0 0] [0] 955.63/297.05 [chk](x1) = [0 0 0] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 2 0] [1] 955.63/297.05 [no](x1) = [0 0 2] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 0 0] [1 0 2] [0] 955.63/297.05 [mat](x1, x2) = [0 0 0] x1 + [0 0 2] x2 + [0] 955.63/297.05 [0 0 0] [0 0 0] [0] 955.63/297.05 955.63/297.05 [0] 955.63/297.05 [X] = [0] 955.63/297.05 [0] 955.63/297.05 955.63/297.05 [0] 955.63/297.05 [y] = [0] 955.63/297.05 [0] 955.63/297.05 955.63/297.05 [0] 955.63/297.05 [c] = [0] 955.63/297.05 [1] 955.63/297.05 955.63/297.05 [1 0 0] [0] 955.63/297.05 [tp](x1) = [0 0 0] x1 + [0] 955.63/297.05 [0 0 0] [1] 955.63/297.05 955.63/297.05 The order satisfies the following ordering constraints: 955.63/297.05 955.63/297.05 [active(f(x))] = [1 1 1] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 1 1] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [mark(f(f(x)))] 955.63/297.05 955.63/297.05 [f(active(x))] = [1 1 1] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 1 1] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [active(f(x))] 955.63/297.05 955.63/297.05 [f(mark(x))] = [1 1 4] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 1 1] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [mark(f(x))] 955.63/297.05 955.63/297.05 [f(no(x))] = [1 2 2] [1] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 1 2] [1] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [no(f(x))] 955.63/297.05 955.63/297.05 [chk(no(f(x)))] = [1 1 2] [1] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 > [1 0 2] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 [chk(no(c()))] = [1] 955.63/297.05 [0] 955.63/297.05 [0] 955.63/297.05 >= [1] 955.63/297.05 [0] 955.63/297.05 [0] 955.63/297.05 = [active(c())] 955.63/297.05 955.63/297.05 [mat(f(x), f(y()))] = [1 1 0] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 0 0] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [f(mat(x, y()))] 955.63/297.05 955.63/297.05 [mat(f(x), c())] = [1 1 0] [2] 955.63/297.05 [0 0 0] x + [2] 955.63/297.05 [0 0 0] [0] 955.63/297.05 > [1] 955.63/297.05 [2] 955.63/297.05 [0] 955.63/297.05 = [no(c())] 955.63/297.05 955.63/297.05 [tp(mark(x))] = [1 1 2] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [1] 955.63/297.05 >= [1 0 2] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [1] 955.63/297.05 = [tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 955.63/297.05 We return to the main proof. 955.63/297.05 955.63/297.05 We are left with following problem, upon which TcT provides the 955.63/297.05 certificate YES(O(1),O(n^2)). 955.63/297.05 955.63/297.05 Strict Trs: 955.63/297.05 { f(active(x)) -> active(f(x)) 955.63/297.05 , f(mark(x)) -> mark(f(x)) 955.63/297.05 , f(no(x)) -> no(f(x)) } 955.63/297.05 Weak Trs: 955.63/297.05 { active(f(x)) -> mark(f(f(x))) 955.63/297.05 , chk(no(f(x))) -> 955.63/297.05 f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) 955.63/297.05 , chk(no(c())) -> active(c()) 955.63/297.05 , mat(f(x), f(y())) -> f(mat(x, y())) 955.63/297.05 , mat(f(x), c()) -> no(c()) 955.63/297.05 , tp(mark(x)) -> 955.63/297.05 tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) } 955.63/297.05 Obligation: 955.63/297.05 derivational complexity 955.63/297.05 Answer: 955.63/297.05 YES(O(1),O(n^2)) 955.63/297.05 955.63/297.05 We use the processor 'matrix interpretation of dimension 2' to 955.63/297.05 orient following rules strictly. 955.63/297.05 955.63/297.05 Trs: { f(no(x)) -> no(f(x)) } 955.63/297.05 955.63/297.05 The induced complexity on above rules (modulo remaining rules) is 955.63/297.05 YES(?,O(n^2)) . These rules are moved into the corresponding weak 955.63/297.05 component(s). 955.63/297.05 955.63/297.05 Sub-proof: 955.63/297.05 ---------- 955.63/297.05 TcT has computed the following triangular matrix interpretation. 955.63/297.05 955.63/297.05 [active](x1) = [1 2] x1 + [0] 955.63/297.05 [0 1] [0] 955.63/297.05 955.63/297.05 [f](x1) = [1 2] x1 + [0] 955.63/297.05 [0 1] [0] 955.63/297.05 955.63/297.05 [mark](x1) = [1 0] x1 + [0] 955.63/297.05 [0 1] [0] 955.63/297.05 955.63/297.05 [chk](x1) = [1 0] x1 + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 955.63/297.05 [no](x1) = [1 0] x1 + [0] 955.63/297.05 [0 1] [2] 955.63/297.05 955.63/297.05 [mat](x1, x2) = [1 0] x1 + [1 0] x2 + [0] 955.63/297.05 [0 0] [0 0] [2] 955.63/297.05 955.63/297.05 [X] = [0] 955.63/297.05 [0] 955.63/297.05 955.63/297.05 [y] = [0] 955.63/297.05 [2] 955.63/297.05 955.63/297.05 [c] = [0] 955.63/297.05 [0] 955.63/297.05 955.63/297.05 [tp](x1) = [1 0] x1 + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 955.63/297.05 The order satisfies the following ordering constraints: 955.63/297.05 955.63/297.05 [active(f(x))] = [1 4] x + [0] 955.63/297.05 [0 1] [0] 955.63/297.05 >= [1 4] x + [0] 955.63/297.05 [0 1] [0] 955.63/297.05 = [mark(f(f(x)))] 955.63/297.05 955.63/297.05 [f(active(x))] = [1 4] x + [0] 955.63/297.05 [0 1] [0] 955.63/297.05 >= [1 4] x + [0] 955.63/297.05 [0 1] [0] 955.63/297.05 = [active(f(x))] 955.63/297.05 955.63/297.05 [f(mark(x))] = [1 2] x + [0] 955.63/297.05 [0 1] [0] 955.63/297.05 >= [1 2] x + [0] 955.63/297.05 [0 1] [0] 955.63/297.05 = [mark(f(x))] 955.63/297.05 955.63/297.05 [f(no(x))] = [1 2] x + [4] 955.63/297.05 [0 1] [2] 955.63/297.05 > [1 2] x + [0] 955.63/297.05 [0 1] [2] 955.63/297.05 = [no(f(x))] 955.63/297.05 955.63/297.05 [chk(no(f(x)))] = [1 2] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 >= [1 0] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 = [f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 [chk(no(c()))] = [0] 955.63/297.05 [0] 955.63/297.05 >= [0] 955.63/297.05 [0] 955.63/297.05 = [active(c())] 955.63/297.05 955.63/297.05 [mat(f(x), f(y()))] = [1 2] x + [4] 955.63/297.05 [0 0] [2] 955.63/297.05 >= [1 0] x + [4] 955.63/297.05 [0 0] [2] 955.63/297.05 = [f(mat(x, y()))] 955.63/297.05 955.63/297.05 [mat(f(x), c())] = [1 2] x + [0] 955.63/297.05 [0 0] [2] 955.63/297.05 >= [0] 955.63/297.05 [2] 955.63/297.05 = [no(c())] 955.63/297.05 955.63/297.05 [tp(mark(x))] = [1 0] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 >= [1 0] x + [0] 955.63/297.05 [0 0] [0] 955.63/297.05 = [tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 955.63/297.05 We return to the main proof. 955.63/297.05 955.63/297.05 We are left with following problem, upon which TcT provides the 955.63/297.05 certificate YES(O(1),O(n^2)). 955.63/297.05 955.63/297.05 Strict Trs: 955.63/297.05 { f(active(x)) -> active(f(x)) 955.63/297.05 , f(mark(x)) -> mark(f(x)) } 955.63/297.05 Weak Trs: 955.63/297.05 { active(f(x)) -> mark(f(f(x))) 955.63/297.05 , f(no(x)) -> no(f(x)) 955.63/297.05 , chk(no(f(x))) -> 955.63/297.05 f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) 955.63/297.05 , chk(no(c())) -> active(c()) 955.63/297.05 , mat(f(x), f(y())) -> f(mat(x, y())) 955.63/297.05 , mat(f(x), c()) -> no(c()) 955.63/297.05 , tp(mark(x)) -> 955.63/297.05 tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) } 955.63/297.05 Obligation: 955.63/297.05 derivational complexity 955.63/297.05 Answer: 955.63/297.05 YES(O(1),O(n^2)) 955.63/297.05 955.63/297.05 We use the processor 'matrix interpretation of dimension 3' to 955.63/297.05 orient following rules strictly. 955.63/297.05 955.63/297.05 Trs: { f(active(x)) -> active(f(x)) } 955.63/297.05 955.63/297.05 The induced complexity on above rules (modulo remaining rules) is 955.63/297.05 YES(?,O(n^2)) . These rules are moved into the corresponding weak 955.63/297.05 component(s). 955.63/297.05 955.63/297.05 Sub-proof: 955.63/297.05 ---------- 955.63/297.05 TcT has computed the following triangular matrix interpretation. 955.63/297.05 Note that the diagonal of the component-wise maxima of 955.63/297.05 interpretation-entries contains no more than 2 non-zero entries. 955.63/297.05 955.63/297.05 [1 1 1] [0] 955.63/297.05 [active](x1) = [0 1 0] x1 + [1] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 1 1] [0] 955.63/297.05 [f](x1) = [0 1 0] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 0 1] [0] 955.63/297.05 [mark](x1) = [0 1 0] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 0 0] [0] 955.63/297.05 [chk](x1) = [0 0 0] x1 + [1] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 0 0] [1] 955.63/297.05 [no](x1) = [0 1 1] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 [1 0 0] [1 0 1] [0] 955.63/297.05 [mat](x1, x2) = [0 0 0] x1 + [0 0 2] x2 + [2] 955.63/297.05 [0 0 0] [0 0 0] [0] 955.63/297.05 955.63/297.05 [0] 955.63/297.05 [X] = [0] 955.63/297.05 [0] 955.63/297.05 955.63/297.05 [2] 955.63/297.05 [y] = [2] 955.63/297.05 [0] 955.63/297.05 955.63/297.05 [0] 955.63/297.05 [c] = [0] 955.63/297.05 [1] 955.63/297.05 955.63/297.05 [1 0 0] [0] 955.63/297.05 [tp](x1) = [0 0 0] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 The order satisfies the following ordering constraints: 955.63/297.05 955.63/297.05 [active(f(x))] = [1 2 1] [0] 955.63/297.05 [0 1 0] x + [1] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 2 1] [0] 955.63/297.05 [0 1 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [mark(f(f(x)))] 955.63/297.05 955.63/297.05 [f(active(x))] = [1 2 1] [1] 955.63/297.05 [0 1 0] x + [1] 955.63/297.05 [0 0 0] [0] 955.63/297.05 > [1 2 1] [0] 955.63/297.05 [0 1 0] x + [1] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [active(f(x))] 955.63/297.05 955.63/297.05 [f(mark(x))] = [1 1 1] [0] 955.63/297.05 [0 1 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 1 1] [0] 955.63/297.05 [0 1 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [mark(f(x))] 955.63/297.05 955.63/297.05 [f(no(x))] = [1 1 1] [1] 955.63/297.05 [0 1 1] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 1 1] [1] 955.63/297.05 [0 1 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [no(f(x))] 955.63/297.05 955.63/297.05 [chk(no(f(x)))] = [1 1 1] [1] 955.63/297.05 [0 0 0] x + [1] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 0 1] [1] 955.63/297.05 [0 0 0] x + [1] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 [chk(no(c()))] = [1] 955.63/297.05 [1] 955.63/297.05 [0] 955.63/297.05 >= [1] 955.63/297.05 [1] 955.63/297.05 [0] 955.63/297.05 = [active(c())] 955.63/297.05 955.63/297.05 [mat(f(x), f(y()))] = [1 1 1] [4] 955.63/297.05 [0 0 0] x + [2] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 0 0] [4] 955.63/297.05 [0 0 0] x + [2] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [f(mat(x, y()))] 955.63/297.05 955.63/297.05 [mat(f(x), c())] = [1 1 1] [1] 955.63/297.05 [0 0 0] x + [4] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1] 955.63/297.05 [1] 955.63/297.05 [0] 955.63/297.05 = [no(c())] 955.63/297.05 955.63/297.05 [tp(mark(x))] = [1 0 1] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 0 1] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 955.63/297.05 We return to the main proof. 955.63/297.05 955.63/297.05 We are left with following problem, upon which TcT provides the 955.63/297.05 certificate YES(O(1),O(n^2)). 955.63/297.05 955.63/297.05 Strict Trs: { f(mark(x)) -> mark(f(x)) } 955.63/297.05 Weak Trs: 955.63/297.05 { active(f(x)) -> mark(f(f(x))) 955.63/297.05 , f(active(x)) -> active(f(x)) 955.63/297.05 , f(no(x)) -> no(f(x)) 955.63/297.05 , chk(no(f(x))) -> 955.63/297.05 f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) 955.63/297.05 , chk(no(c())) -> active(c()) 955.63/297.05 , mat(f(x), f(y())) -> f(mat(x, y())) 955.63/297.05 , mat(f(x), c()) -> no(c()) 955.63/297.05 , tp(mark(x)) -> 955.63/297.05 tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) } 955.63/297.05 Obligation: 955.63/297.05 derivational complexity 955.63/297.05 Answer: 955.63/297.05 YES(O(1),O(n^2)) 955.63/297.05 955.63/297.05 We use the processor 'matrix interpretation of dimension 3' to 955.63/297.05 orient following rules strictly. 955.63/297.05 955.63/297.05 Trs: { f(mark(x)) -> mark(f(x)) } 955.63/297.05 955.63/297.05 The induced complexity on above rules (modulo remaining rules) is 955.63/297.05 YES(?,O(n^2)) . These rules are moved into the corresponding weak 955.63/297.05 component(s). 955.63/297.05 955.63/297.05 Sub-proof: 955.63/297.05 ---------- 955.63/297.05 TcT has computed the following triangular matrix interpretation. 955.63/297.05 Note that the diagonal of the component-wise maxima of 955.63/297.05 interpretation-entries contains no more than 2 non-zero entries. 955.63/297.05 955.63/297.05 [1 0 2] [0] 955.63/297.05 [active](x1) = [0 0 0] x1 + [0] 955.63/297.05 [0 0 1] [2] 955.63/297.05 955.63/297.05 [1 0 2] [0] 955.63/297.05 [f](x1) = [0 0 0] x1 + [2] 955.63/297.05 [0 0 1] [0] 955.63/297.05 955.63/297.05 [1 0 0] [0] 955.63/297.05 [mark](x1) = [0 0 0] x1 + [0] 955.63/297.05 [0 0 1] [2] 955.63/297.05 955.63/297.05 [1 0 0] [0] 955.63/297.05 [chk](x1) = [0 0 0] x1 + [2] 955.63/297.05 [0 0 0] [2] 955.63/297.05 955.63/297.05 [1 2 0] [0] 955.63/297.05 [no](x1) = [0 0 0] x1 + [0] 955.63/297.05 [0 0 1] [2] 955.63/297.05 955.63/297.05 [1 0 0] [1 0 0] [0] 955.63/297.05 [mat](x1, x2) = [0 0 0] x1 + [0 0 2] x2 + [0] 955.63/297.05 [0 0 0] [0 0 0] [2] 955.63/297.05 955.63/297.05 [0] 955.63/297.05 [X] = [0] 955.63/297.05 [0] 955.63/297.05 955.63/297.05 [0] 955.63/297.05 [y] = [0] 955.63/297.05 [2] 955.63/297.05 955.63/297.05 [0] 955.63/297.05 [c] = [0] 955.63/297.05 [0] 955.63/297.05 955.63/297.05 [1 0 0] [0] 955.63/297.05 [tp](x1) = [0 0 0] x1 + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 955.63/297.05 The order satisfies the following ordering constraints: 955.63/297.05 955.63/297.05 [active(f(x))] = [1 0 4] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 1] [2] 955.63/297.05 >= [1 0 4] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 1] [2] 955.63/297.05 = [mark(f(f(x)))] 955.63/297.05 955.63/297.05 [f(active(x))] = [1 0 4] [4] 955.63/297.05 [0 0 0] x + [2] 955.63/297.05 [0 0 1] [2] 955.63/297.05 > [1 0 4] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 1] [2] 955.63/297.05 = [active(f(x))] 955.63/297.05 955.63/297.05 [f(mark(x))] = [1 0 2] [4] 955.63/297.05 [0 0 0] x + [2] 955.63/297.05 [0 0 1] [2] 955.63/297.05 > [1 0 2] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 1] [2] 955.63/297.05 = [mark(f(x))] 955.63/297.05 955.63/297.05 [f(no(x))] = [1 2 2] [4] 955.63/297.05 [0 0 0] x + [2] 955.63/297.05 [0 0 1] [2] 955.63/297.05 >= [1 0 2] [4] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 1] [2] 955.63/297.05 = [no(f(x))] 955.63/297.05 955.63/297.05 [chk(no(f(x)))] = [1 0 2] [4] 955.63/297.05 [0 0 0] x + [2] 955.63/297.05 [0 0 0] [2] 955.63/297.05 >= [1 0 0] [4] 955.63/297.05 [0 0 0] x + [2] 955.63/297.05 [0 0 0] [2] 955.63/297.05 = [f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 [chk(no(c()))] = [0] 955.63/297.05 [2] 955.63/297.05 [2] 955.63/297.05 >= [0] 955.63/297.05 [0] 955.63/297.05 [2] 955.63/297.05 = [active(c())] 955.63/297.05 955.63/297.05 [mat(f(x), f(y()))] = [1 0 2] [4] 955.63/297.05 [0 0 0] x + [4] 955.63/297.05 [0 0 0] [2] 955.63/297.05 >= [1 0 0] [4] 955.63/297.05 [0 0 0] x + [2] 955.63/297.05 [0 0 0] [2] 955.63/297.05 = [f(mat(x, y()))] 955.63/297.05 955.63/297.05 [mat(f(x), c())] = [1 0 2] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [2] 955.63/297.05 >= [0] 955.63/297.05 [0] 955.63/297.05 [2] 955.63/297.05 = [no(c())] 955.63/297.05 955.63/297.05 [tp(mark(x))] = [1 0 0] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 >= [1 0 0] [0] 955.63/297.05 [0 0 0] x + [0] 955.63/297.05 [0 0 0] [0] 955.63/297.05 = [tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))] 955.63/297.05 955.63/297.05 955.63/297.05 We return to the main proof. 955.63/297.05 955.63/297.05 We are left with following problem, upon which TcT provides the 955.63/297.05 certificate YES(O(1),O(1)). 955.63/297.05 955.63/297.05 Weak Trs: 955.63/297.05 { active(f(x)) -> mark(f(f(x))) 955.63/297.05 , f(active(x)) -> active(f(x)) 955.63/297.05 , f(mark(x)) -> mark(f(x)) 955.63/297.05 , f(no(x)) -> no(f(x)) 955.63/297.05 , chk(no(f(x))) -> 955.63/297.05 f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) 955.63/297.05 , chk(no(c())) -> active(c()) 955.63/297.05 , mat(f(x), f(y())) -> f(mat(x, y())) 955.63/297.05 , mat(f(x), c()) -> no(c()) 955.63/297.05 , tp(mark(x)) -> 955.63/297.05 tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x))) } 955.63/297.05 Obligation: 955.63/297.05 derivational complexity 955.63/297.05 Answer: 955.63/297.05 YES(O(1),O(1)) 955.63/297.05 955.63/297.05 Empty rules are trivially bounded 955.63/297.05 955.63/297.05 Hurray, we answered YES(O(1),O(n^2)) 956.08/297.45 EOF