MAYBE 201.10/60.05 MAYBE 201.10/60.05 201.10/60.05 We are left with following problem, upon which TcT provides the 201.10/60.05 certificate MAYBE. 201.10/60.05 201.10/60.05 Strict Trs: 201.10/60.05 { double(0()) -> 0() 201.10/60.05 , double(s(x)) -> s(s(double(x))) 201.10/60.05 , half(double(x)) -> x 201.10/60.05 , half(0()) -> 0() 201.10/60.05 , half(s(0())) -> 0() 201.10/60.05 , half(s(s(x))) -> s(half(x)) 201.10/60.05 , -(x, 0()) -> x 201.10/60.05 , -(s(x), s(y)) -> -(x, y) 201.10/60.05 , if(0(), y, z) -> y 201.10/60.05 , if(s(x), y, z) -> z } 201.10/60.05 Obligation: 201.10/60.05 derivational complexity 201.10/60.05 Answer: 201.10/60.05 MAYBE 201.10/60.05 201.10/60.05 None of the processors succeeded. 201.10/60.05 201.10/60.05 Details of failed attempt(s): 201.10/60.05 ----------------------------- 201.10/60.05 1) 'Fastest (timeout of 60 seconds)' failed due to the following 201.10/60.05 reason: 201.10/60.05 201.10/60.05 Computation stopped due to timeout after 60.0 seconds. 201.10/60.05 201.10/60.05 2) 'Inspecting Problem... (timeout of 297 seconds)' failed due to 201.10/60.05 the following reason: 201.10/60.05 201.10/60.05 We use the processor 'matrix interpretation of dimension 1' to 201.10/60.05 orient following rules strictly. 201.10/60.05 201.10/60.05 Trs: 201.10/60.05 { if(0(), y, z) -> y 201.10/60.05 , if(s(x), y, z) -> z } 201.10/60.05 201.10/60.05 The induced complexity on above rules (modulo remaining rules) is 201.10/60.05 YES(?,O(n^1)) . These rules are moved into the corresponding weak 201.10/60.05 component(s). 201.10/60.05 201.10/60.05 Sub-proof: 201.10/60.05 ---------- 201.10/60.05 TcT has computed the following triangular matrix interpretation. 201.10/60.05 201.10/60.05 [double](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [0] = [0] 201.10/60.05 201.10/60.05 [s](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [half](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [-](x1, x2) = [1] x1 + [1] x2 + [0] 201.10/60.05 201.10/60.05 [if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1] 201.10/60.05 201.10/60.05 The order satisfies the following ordering constraints: 201.10/60.05 201.10/60.05 [double(0())] = [0] 201.10/60.05 >= [0] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [double(s(x))] = [1] x + [0] 201.10/60.05 >= [1] x + [0] 201.10/60.05 = [s(s(double(x)))] 201.10/60.05 201.10/60.05 [half(double(x))] = [1] x + [0] 201.10/60.05 >= [1] x + [0] 201.10/60.05 = [x] 201.10/60.05 201.10/60.05 [half(0())] = [0] 201.10/60.05 >= [0] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [half(s(0()))] = [0] 201.10/60.05 >= [0] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [half(s(s(x)))] = [1] x + [0] 201.10/60.05 >= [1] x + [0] 201.10/60.05 = [s(half(x))] 201.10/60.05 201.10/60.05 [-(x, 0())] = [1] x + [0] 201.10/60.05 >= [1] x + [0] 201.10/60.05 = [x] 201.10/60.05 201.10/60.05 [-(s(x), s(y))] = [1] x + [1] y + [0] 201.10/60.05 >= [1] x + [1] y + [0] 201.10/60.05 = [-(x, y)] 201.10/60.05 201.10/60.05 [if(0(), y, z)] = [1] y + [1] z + [1] 201.10/60.05 > [1] y + [0] 201.10/60.05 = [y] 201.10/60.05 201.10/60.05 [if(s(x), y, z)] = [1] x + [1] y + [1] z + [1] 201.10/60.05 > [1] z + [0] 201.10/60.05 = [z] 201.10/60.05 201.10/60.05 201.10/60.05 We return to the main proof. 201.10/60.05 201.10/60.05 We are left with following problem, upon which TcT provides the 201.10/60.05 certificate MAYBE. 201.10/60.05 201.10/60.05 Strict Trs: 201.10/60.05 { double(0()) -> 0() 201.10/60.05 , double(s(x)) -> s(s(double(x))) 201.10/60.05 , half(double(x)) -> x 201.10/60.05 , half(0()) -> 0() 201.10/60.05 , half(s(0())) -> 0() 201.10/60.05 , half(s(s(x))) -> s(half(x)) 201.10/60.05 , -(x, 0()) -> x 201.10/60.05 , -(s(x), s(y)) -> -(x, y) } 201.10/60.05 Weak Trs: 201.10/60.05 { if(0(), y, z) -> y 201.10/60.05 , if(s(x), y, z) -> z } 201.10/60.05 Obligation: 201.10/60.05 derivational complexity 201.10/60.05 Answer: 201.10/60.05 MAYBE 201.10/60.05 201.10/60.05 We use the processor 'matrix interpretation of dimension 1' to 201.10/60.05 orient following rules strictly. 201.10/60.05 201.10/60.05 Trs: { -(x, 0()) -> x } 201.10/60.05 201.10/60.05 The induced complexity on above rules (modulo remaining rules) is 201.10/60.05 YES(?,O(n^1)) . These rules are moved into the corresponding weak 201.10/60.05 component(s). 201.10/60.05 201.10/60.05 Sub-proof: 201.10/60.05 ---------- 201.10/60.05 TcT has computed the following triangular matrix interpretation. 201.10/60.05 201.10/60.05 [double](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [0] = [1] 201.10/60.05 201.10/60.05 [s](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [half](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [-](x1, x2) = [1] x1 + [1] x2 + [0] 201.10/60.05 201.10/60.05 [if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1] 201.10/60.05 201.10/60.05 The order satisfies the following ordering constraints: 201.10/60.05 201.10/60.05 [double(0())] = [1] 201.10/60.05 >= [1] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [double(s(x))] = [1] x + [0] 201.10/60.05 >= [1] x + [0] 201.10/60.05 = [s(s(double(x)))] 201.10/60.05 201.10/60.05 [half(double(x))] = [1] x + [0] 201.10/60.05 >= [1] x + [0] 201.10/60.05 = [x] 201.10/60.05 201.10/60.05 [half(0())] = [1] 201.10/60.05 >= [1] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [half(s(0()))] = [1] 201.10/60.05 >= [1] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [half(s(s(x)))] = [1] x + [0] 201.10/60.05 >= [1] x + [0] 201.10/60.05 = [s(half(x))] 201.10/60.05 201.10/60.05 [-(x, 0())] = [1] x + [1] 201.10/60.05 > [1] x + [0] 201.10/60.05 = [x] 201.10/60.05 201.10/60.05 [-(s(x), s(y))] = [1] x + [1] y + [0] 201.10/60.05 >= [1] x + [1] y + [0] 201.10/60.05 = [-(x, y)] 201.10/60.05 201.10/60.05 [if(0(), y, z)] = [1] y + [1] z + [2] 201.10/60.05 > [1] y + [0] 201.10/60.05 = [y] 201.10/60.05 201.10/60.05 [if(s(x), y, z)] = [1] x + [1] y + [1] z + [1] 201.10/60.05 > [1] z + [0] 201.10/60.05 = [z] 201.10/60.05 201.10/60.05 201.10/60.05 We return to the main proof. 201.10/60.05 201.10/60.05 We are left with following problem, upon which TcT provides the 201.10/60.05 certificate MAYBE. 201.10/60.05 201.10/60.05 Strict Trs: 201.10/60.05 { double(0()) -> 0() 201.10/60.05 , double(s(x)) -> s(s(double(x))) 201.10/60.05 , half(double(x)) -> x 201.10/60.05 , half(0()) -> 0() 201.10/60.05 , half(s(0())) -> 0() 201.10/60.05 , half(s(s(x))) -> s(half(x)) 201.10/60.05 , -(s(x), s(y)) -> -(x, y) } 201.10/60.05 Weak Trs: 201.10/60.05 { -(x, 0()) -> x 201.10/60.05 , if(0(), y, z) -> y 201.10/60.05 , if(s(x), y, z) -> z } 201.10/60.05 Obligation: 201.10/60.05 derivational complexity 201.10/60.05 Answer: 201.10/60.05 MAYBE 201.10/60.05 201.10/60.05 The weightgap principle applies (using the following nonconstant 201.10/60.05 growth matrix-interpretation) 201.10/60.05 201.10/60.05 TcT has computed the following triangular matrix interpretation. 201.10/60.05 Note that the diagonal of the component-wise maxima of 201.10/60.05 interpretation-entries contains no more than 1 non-zero entries. 201.10/60.05 201.10/60.05 [double](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [0] = [0] 201.10/60.05 201.10/60.05 [s](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [half](x1) = [1] x1 + [1] 201.10/60.05 201.10/60.05 [-](x1, x2) = [1] x1 + [1] x2 + [2] 201.10/60.05 201.10/60.05 [if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [2] 201.10/60.05 201.10/60.05 The order satisfies the following ordering constraints: 201.10/60.05 201.10/60.05 [double(0())] = [0] 201.10/60.05 >= [0] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [double(s(x))] = [1] x + [0] 201.10/60.05 >= [1] x + [0] 201.10/60.05 = [s(s(double(x)))] 201.10/60.05 201.10/60.05 [half(double(x))] = [1] x + [1] 201.10/60.05 > [1] x + [0] 201.10/60.05 = [x] 201.10/60.05 201.10/60.05 [half(0())] = [1] 201.10/60.05 > [0] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [half(s(0()))] = [1] 201.10/60.05 > [0] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [half(s(s(x)))] = [1] x + [1] 201.10/60.05 >= [1] x + [1] 201.10/60.05 = [s(half(x))] 201.10/60.05 201.10/60.05 [-(x, 0())] = [1] x + [2] 201.10/60.05 > [1] x + [0] 201.10/60.05 = [x] 201.10/60.05 201.10/60.05 [-(s(x), s(y))] = [1] x + [1] y + [2] 201.10/60.05 >= [1] x + [1] y + [2] 201.10/60.05 = [-(x, y)] 201.10/60.05 201.10/60.05 [if(0(), y, z)] = [1] y + [1] z + [2] 201.10/60.05 > [1] y + [0] 201.10/60.05 = [y] 201.10/60.05 201.10/60.05 [if(s(x), y, z)] = [1] x + [1] y + [1] z + [2] 201.10/60.05 > [1] z + [0] 201.10/60.05 = [z] 201.10/60.05 201.10/60.05 201.10/60.05 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 201.10/60.05 201.10/60.05 We are left with following problem, upon which TcT provides the 201.10/60.05 certificate MAYBE. 201.10/60.05 201.10/60.05 Strict Trs: 201.10/60.05 { double(0()) -> 0() 201.10/60.05 , double(s(x)) -> s(s(double(x))) 201.10/60.05 , half(s(s(x))) -> s(half(x)) 201.10/60.05 , -(s(x), s(y)) -> -(x, y) } 201.10/60.05 Weak Trs: 201.10/60.05 { half(double(x)) -> x 201.10/60.05 , half(0()) -> 0() 201.10/60.05 , half(s(0())) -> 0() 201.10/60.05 , -(x, 0()) -> x 201.10/60.05 , if(0(), y, z) -> y 201.10/60.05 , if(s(x), y, z) -> z } 201.10/60.05 Obligation: 201.10/60.05 derivational complexity 201.10/60.05 Answer: 201.10/60.05 MAYBE 201.10/60.05 201.10/60.05 We use the processor 'matrix interpretation of dimension 1' to 201.10/60.05 orient following rules strictly. 201.10/60.05 201.10/60.05 Trs: { double(0()) -> 0() } 201.10/60.05 201.10/60.05 The induced complexity on above rules (modulo remaining rules) is 201.10/60.05 YES(?,O(n^1)) . These rules are moved into the corresponding weak 201.10/60.05 component(s). 201.10/60.05 201.10/60.05 Sub-proof: 201.10/60.05 ---------- 201.10/60.05 TcT has computed the following triangular matrix interpretation. 201.10/60.05 201.10/60.05 [double](x1) = [1] x1 + [1] 201.10/60.05 201.10/60.05 [0] = [0] 201.10/60.05 201.10/60.05 [s](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [half](x1) = [1] x1 + [0] 201.10/60.05 201.10/60.05 [-](x1, x2) = [1] x1 + [1] x2 + [2] 201.10/60.05 201.10/60.05 [if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [2] 201.10/60.05 201.10/60.05 The order satisfies the following ordering constraints: 201.10/60.05 201.10/60.05 [double(0())] = [1] 201.10/60.05 > [0] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [double(s(x))] = [1] x + [1] 201.10/60.05 >= [1] x + [1] 201.10/60.05 = [s(s(double(x)))] 201.10/60.05 201.10/60.05 [half(double(x))] = [1] x + [1] 201.10/60.05 > [1] x + [0] 201.10/60.05 = [x] 201.10/60.05 201.10/60.05 [half(0())] = [0] 201.10/60.05 >= [0] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [half(s(0()))] = [0] 201.10/60.05 >= [0] 201.10/60.05 = [0()] 201.10/60.05 201.10/60.05 [half(s(s(x)))] = [1] x + [0] 201.10/60.05 >= [1] x + [0] 201.10/60.05 = [s(half(x))] 201.10/60.05 201.10/60.05 [-(x, 0())] = [1] x + [2] 201.10/60.05 > [1] x + [0] 201.10/60.05 = [x] 201.10/60.05 201.10/60.05 [-(s(x), s(y))] = [1] x + [1] y + [2] 201.10/60.05 >= [1] x + [1] y + [2] 201.10/60.05 = [-(x, y)] 201.10/60.05 201.10/60.05 [if(0(), y, z)] = [1] y + [1] z + [2] 201.10/60.05 > [1] y + [0] 201.10/60.05 = [y] 201.10/60.05 201.10/60.05 [if(s(x), y, z)] = [1] x + [1] y + [1] z + [2] 201.10/60.05 > [1] z + [0] 201.10/60.05 = [z] 201.10/60.05 201.10/60.05 201.10/60.05 We return to the main proof. 201.10/60.05 201.10/60.05 We are left with following problem, upon which TcT provides the 201.10/60.05 certificate MAYBE. 201.10/60.05 201.10/60.05 Strict Trs: 201.10/60.05 { double(s(x)) -> s(s(double(x))) 201.10/60.05 , half(s(s(x))) -> s(half(x)) 201.10/60.05 , -(s(x), s(y)) -> -(x, y) } 201.10/60.05 Weak Trs: 201.10/60.05 { double(0()) -> 0() 201.10/60.05 , half(double(x)) -> x 201.10/60.05 , half(0()) -> 0() 201.10/60.05 , half(s(0())) -> 0() 201.10/60.05 , -(x, 0()) -> x 201.10/60.05 , if(0(), y, z) -> y 201.10/60.05 , if(s(x), y, z) -> z } 201.10/60.05 Obligation: 201.10/60.05 derivational complexity 201.10/60.05 Answer: 201.10/60.05 MAYBE 201.10/60.05 201.10/60.05 The weightgap principle applies (using the following nonconstant 201.10/60.05 growth matrix-interpretation) 201.10/60.06 201.10/60.06 TcT has computed the following triangular matrix interpretation. 201.10/60.06 Note that the diagonal of the component-wise maxima of 201.10/60.06 interpretation-entries contains no more than 1 non-zero entries. 201.10/60.06 201.10/60.06 [double](x1) = [1] x1 + [0] 201.10/60.06 201.10/60.06 [0] = [2] 201.10/60.06 201.10/60.06 [s](x1) = [1] x1 + [1] 201.10/60.06 201.10/60.06 [half](x1) = [1] x1 + [2] 201.10/60.06 201.10/60.06 [-](x1, x2) = [1] x1 + [1] x2 + [1] 201.10/60.06 201.10/60.06 [if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1] 201.10/60.06 201.10/60.06 The order satisfies the following ordering constraints: 201.10/60.06 201.10/60.06 [double(0())] = [2] 201.10/60.06 >= [2] 201.10/60.06 = [0()] 201.10/60.06 201.10/60.06 [double(s(x))] = [1] x + [1] 201.10/60.06 ? [1] x + [2] 201.10/60.06 = [s(s(double(x)))] 201.10/60.06 201.10/60.06 [half(double(x))] = [1] x + [2] 201.10/60.06 > [1] x + [0] 201.10/60.06 = [x] 201.10/60.06 201.10/60.06 [half(0())] = [4] 201.10/60.06 > [2] 201.10/60.06 = [0()] 201.10/60.06 201.10/60.06 [half(s(0()))] = [5] 201.10/60.06 > [2] 201.10/60.06 = [0()] 201.10/60.06 201.10/60.06 [half(s(s(x)))] = [1] x + [4] 201.10/60.06 > [1] x + [3] 201.10/60.06 = [s(half(x))] 201.10/60.06 201.10/60.06 [-(x, 0())] = [1] x + [3] 201.10/60.06 > [1] x + [0] 201.10/60.06 = [x] 201.10/60.06 201.10/60.06 [-(s(x), s(y))] = [1] x + [1] y + [3] 201.10/60.06 > [1] x + [1] y + [1] 201.10/60.06 = [-(x, y)] 201.10/60.06 201.10/60.06 [if(0(), y, z)] = [1] y + [1] z + [3] 201.10/60.06 > [1] y + [0] 201.10/60.06 = [y] 201.10/60.06 201.10/60.06 [if(s(x), y, z)] = [1] x + [1] y + [1] z + [2] 201.10/60.06 > [1] z + [0] 201.10/60.06 = [z] 201.10/60.06 201.10/60.06 201.10/60.06 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 201.10/60.06 201.10/60.06 We are left with following problem, upon which TcT provides the 201.10/60.06 certificate MAYBE. 201.10/60.06 201.10/60.06 Strict Trs: { double(s(x)) -> s(s(double(x))) } 201.10/60.06 Weak Trs: 201.10/60.06 { double(0()) -> 0() 201.10/60.06 , half(double(x)) -> x 201.10/60.06 , half(0()) -> 0() 201.10/60.06 , half(s(0())) -> 0() 201.10/60.06 , half(s(s(x))) -> s(half(x)) 201.10/60.06 , -(x, 0()) -> x 201.10/60.06 , -(s(x), s(y)) -> -(x, y) 201.10/60.06 , if(0(), y, z) -> y 201.10/60.06 , if(s(x), y, z) -> z } 201.10/60.06 Obligation: 201.10/60.06 derivational complexity 201.10/60.06 Answer: 201.10/60.06 MAYBE 201.10/60.06 201.10/60.06 None of the processors succeeded. 201.10/60.06 201.10/60.06 Details of failed attempt(s): 201.10/60.06 ----------------------------- 201.10/60.06 1) 'empty' failed due to the following reason: 201.10/60.06 201.10/60.06 Empty strict component of the problem is NOT empty. 201.10/60.06 201.10/60.06 2) 'Fastest' failed due to the following reason: 201.10/60.06 201.10/60.06 None of the processors succeeded. 201.10/60.06 201.10/60.06 Details of failed attempt(s): 201.10/60.06 ----------------------------- 201.10/60.06 1) 'Fastest (timeout of 30 seconds)' failed due to the following 201.10/60.06 reason: 201.10/60.06 201.10/60.06 Computation stopped due to timeout after 30.0 seconds. 201.10/60.06 201.10/60.06 2) 'Fastest' failed due to the following reason: 201.10/60.06 201.10/60.06 None of the processors succeeded. 201.10/60.06 201.10/60.06 Details of failed attempt(s): 201.10/60.06 ----------------------------- 201.10/60.06 1) 'matrix interpretation of dimension 6' failed due to the 201.10/60.06 following reason: 201.10/60.06 201.10/60.06 The input cannot be shown compatible 201.10/60.06 201.10/60.06 2) 'matrix interpretation of dimension 5' failed due to the 201.10/60.06 following reason: 201.10/60.06 201.10/60.06 The input cannot be shown compatible 201.10/60.06 201.10/60.06 3) 'matrix interpretation of dimension 4' failed due to the 201.10/60.06 following reason: 201.10/60.06 201.10/60.06 The input cannot be shown compatible 201.10/60.06 201.10/60.06 4) 'matrix interpretation of dimension 3' failed due to the 201.10/60.06 following reason: 201.10/60.06 201.10/60.06 The input cannot be shown compatible 201.10/60.06 201.10/60.06 5) 'matrix interpretation of dimension 2' failed due to the 201.10/60.06 following reason: 201.10/60.06 201.10/60.06 The input cannot be shown compatible 201.10/60.06 201.10/60.06 6) 'matrix interpretation of dimension 1' failed due to the 201.10/60.06 following reason: 201.10/60.06 201.10/60.06 The input cannot be shown compatible 201.10/60.06 201.10/60.06 201.10/60.06 3) 'iteProgress' failed due to the following reason: 201.10/60.06 201.10/60.06 Fail 201.10/60.06 201.10/60.06 4) 'bsearch-matrix' failed due to the following reason: 201.10/60.06 201.10/60.06 The input cannot be shown compatible 201.10/60.06 201.10/60.06 201.10/60.06 201.10/60.06 3) 'iteProgress (timeout of 297 seconds)' failed due to the 201.10/60.06 following reason: 201.10/60.06 201.10/60.06 Fail 201.10/60.06 201.10/60.06 4) 'bsearch-matrix (timeout of 297 seconds)' failed due to the 201.10/60.06 following reason: 201.10/60.06 201.10/60.06 The input cannot be shown compatible 201.10/60.06 201.10/60.06 201.10/60.06 Arrrr.. 201.10/60.06 EOF