YES(O(1),O(n^2)) 237.91/60.05 YES(O(1),O(n^2)) 237.91/60.05 237.91/60.05 We are left with following problem, upon which TcT provides the 237.91/60.05 certificate YES(O(1),O(n^2)). 237.91/60.05 237.91/60.05 Strict Trs: 237.91/60.05 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.05 , a(A(x1)) -> x1 237.91/60.05 , b(a(C(x1))) -> C(a(b(x1))) 237.91/60.05 , b(B(x1)) -> x1 237.91/60.05 , c(C(x1)) -> x1 237.91/60.05 , c(A(B(x1))) -> B(A(c(x1))) 237.91/60.05 , C(c(x1)) -> x1 237.91/60.05 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.05 , B(b(x1)) -> x1 237.91/60.05 , B(C(a(x1))) -> a(C(B(x1))) 237.91/60.05 , A(a(x1)) -> x1 237.91/60.05 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.05 Obligation: 237.91/60.05 derivational complexity 237.91/60.05 Answer: 237.91/60.05 YES(O(1),O(n^2)) 237.91/60.05 237.91/60.05 We use the processor 'matrix interpretation of dimension 1' to 237.91/60.05 orient following rules strictly. 237.91/60.05 237.91/60.05 Trs: 237.91/60.05 { a(A(x1)) -> x1 237.91/60.05 , A(a(x1)) -> x1 } 237.91/60.05 237.91/60.05 The induced complexity on above rules (modulo remaining rules) is 237.91/60.05 YES(?,O(n^1)) . These rules are moved into the corresponding weak 237.91/60.05 component(s). 237.91/60.05 237.91/60.05 Sub-proof: 237.91/60.05 ---------- 237.91/60.05 TcT has computed the following triangular matrix interpretation. 237.91/60.05 237.91/60.05 [a](x1) = [1] x1 + [1] 237.91/60.05 237.91/60.05 [b](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [c](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [C](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [B](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [A](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 The order satisfies the following ordering constraints: 237.91/60.05 237.91/60.05 [a(b(c(x1)))] = [1] x1 + [1] 237.91/60.05 >= [1] x1 + [1] 237.91/60.05 = [c(b(a(x1)))] 237.91/60.05 237.91/60.05 [a(A(x1))] = [1] x1 + [1] 237.91/60.05 > [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [b(a(C(x1)))] = [1] x1 + [1] 237.91/60.05 >= [1] x1 + [1] 237.91/60.05 = [C(a(b(x1)))] 237.91/60.05 237.91/60.05 [b(B(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(C(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(A(B(x1)))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [B(A(c(x1)))] 237.91/60.05 237.91/60.05 [C(c(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [C(B(A(x1)))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [A(B(C(x1)))] 237.91/60.05 237.91/60.05 [B(b(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [B(C(a(x1)))] = [1] x1 + [1] 237.91/60.05 >= [1] x1 + [1] 237.91/60.05 = [a(C(B(x1)))] 237.91/60.05 237.91/60.05 [A(a(x1))] = [1] x1 + [1] 237.91/60.05 > [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [A(c(b(x1)))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [b(c(A(x1)))] 237.91/60.05 237.91/60.05 237.91/60.05 We return to the main proof. 237.91/60.05 237.91/60.05 We are left with following problem, upon which TcT provides the 237.91/60.05 certificate YES(O(1),O(n^2)). 237.91/60.05 237.91/60.05 Strict Trs: 237.91/60.05 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.05 , b(a(C(x1))) -> C(a(b(x1))) 237.91/60.05 , b(B(x1)) -> x1 237.91/60.05 , c(C(x1)) -> x1 237.91/60.05 , c(A(B(x1))) -> B(A(c(x1))) 237.91/60.05 , C(c(x1)) -> x1 237.91/60.05 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.05 , B(b(x1)) -> x1 237.91/60.05 , B(C(a(x1))) -> a(C(B(x1))) 237.91/60.05 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.05 Weak Trs: 237.91/60.05 { a(A(x1)) -> x1 237.91/60.05 , A(a(x1)) -> x1 } 237.91/60.05 Obligation: 237.91/60.05 derivational complexity 237.91/60.05 Answer: 237.91/60.05 YES(O(1),O(n^2)) 237.91/60.05 237.91/60.05 We use the processor 'matrix interpretation of dimension 1' to 237.91/60.05 orient following rules strictly. 237.91/60.05 237.91/60.05 Trs: 237.91/60.05 { c(C(x1)) -> x1 237.91/60.05 , C(c(x1)) -> x1 } 237.91/60.05 237.91/60.05 The induced complexity on above rules (modulo remaining rules) is 237.91/60.05 YES(?,O(n^1)) . These rules are moved into the corresponding weak 237.91/60.05 component(s). 237.91/60.05 237.91/60.05 Sub-proof: 237.91/60.05 ---------- 237.91/60.05 TcT has computed the following triangular matrix interpretation. 237.91/60.05 237.91/60.05 [a](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [b](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [c](x1) = [1] x1 + [1] 237.91/60.05 237.91/60.05 [C](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [B](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [A](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 The order satisfies the following ordering constraints: 237.91/60.05 237.91/60.05 [a(b(c(x1)))] = [1] x1 + [1] 237.91/60.05 >= [1] x1 + [1] 237.91/60.05 = [c(b(a(x1)))] 237.91/60.05 237.91/60.05 [a(A(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [b(a(C(x1)))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [C(a(b(x1)))] 237.91/60.05 237.91/60.05 [b(B(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(C(x1))] = [1] x1 + [1] 237.91/60.05 > [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(A(B(x1)))] = [1] x1 + [1] 237.91/60.05 >= [1] x1 + [1] 237.91/60.05 = [B(A(c(x1)))] 237.91/60.05 237.91/60.05 [C(c(x1))] = [1] x1 + [1] 237.91/60.05 > [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [C(B(A(x1)))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [A(B(C(x1)))] 237.91/60.05 237.91/60.05 [B(b(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [B(C(a(x1)))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [a(C(B(x1)))] 237.91/60.05 237.91/60.05 [A(a(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [A(c(b(x1)))] = [1] x1 + [1] 237.91/60.05 >= [1] x1 + [1] 237.91/60.05 = [b(c(A(x1)))] 237.91/60.05 237.91/60.05 237.91/60.05 We return to the main proof. 237.91/60.05 237.91/60.05 We are left with following problem, upon which TcT provides the 237.91/60.05 certificate YES(O(1),O(n^2)). 237.91/60.05 237.91/60.05 Strict Trs: 237.91/60.05 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.05 , b(a(C(x1))) -> C(a(b(x1))) 237.91/60.05 , b(B(x1)) -> x1 237.91/60.05 , c(A(B(x1))) -> B(A(c(x1))) 237.91/60.05 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.05 , B(b(x1)) -> x1 237.91/60.05 , B(C(a(x1))) -> a(C(B(x1))) 237.91/60.05 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.05 Weak Trs: 237.91/60.05 { a(A(x1)) -> x1 237.91/60.05 , c(C(x1)) -> x1 237.91/60.05 , C(c(x1)) -> x1 237.91/60.05 , A(a(x1)) -> x1 } 237.91/60.05 Obligation: 237.91/60.05 derivational complexity 237.91/60.05 Answer: 237.91/60.05 YES(O(1),O(n^2)) 237.91/60.05 237.91/60.05 We use the processor 'matrix interpretation of dimension 1' to 237.91/60.05 orient following rules strictly. 237.91/60.05 237.91/60.05 Trs: 237.91/60.05 { b(B(x1)) -> x1 237.91/60.05 , B(b(x1)) -> x1 } 237.91/60.05 237.91/60.05 The induced complexity on above rules (modulo remaining rules) is 237.91/60.05 YES(?,O(n^1)) . These rules are moved into the corresponding weak 237.91/60.05 component(s). 237.91/60.05 237.91/60.05 Sub-proof: 237.91/60.05 ---------- 237.91/60.05 TcT has computed the following triangular matrix interpretation. 237.91/60.05 237.91/60.05 [a](x1) = [1] x1 + [2] 237.91/60.05 237.91/60.05 [b](x1) = [1] x1 + [2] 237.91/60.05 237.91/60.05 [c](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [C](x1) = [1] x1 + [0] 237.91/60.05 237.91/60.05 [B](x1) = [1] x1 + [2] 237.91/60.05 237.91/60.05 [A](x1) = [1] x1 + [2] 237.91/60.05 237.91/60.05 The order satisfies the following ordering constraints: 237.91/60.05 237.91/60.05 [a(b(c(x1)))] = [1] x1 + [4] 237.91/60.05 >= [1] x1 + [4] 237.91/60.05 = [c(b(a(x1)))] 237.91/60.05 237.91/60.05 [a(A(x1))] = [1] x1 + [4] 237.91/60.05 > [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [b(a(C(x1)))] = [1] x1 + [4] 237.91/60.05 >= [1] x1 + [4] 237.91/60.05 = [C(a(b(x1)))] 237.91/60.05 237.91/60.05 [b(B(x1))] = [1] x1 + [4] 237.91/60.05 > [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(C(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(A(B(x1)))] = [1] x1 + [4] 237.91/60.05 >= [1] x1 + [4] 237.91/60.05 = [B(A(c(x1)))] 237.91/60.05 237.91/60.05 [C(c(x1))] = [1] x1 + [0] 237.91/60.05 >= [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [C(B(A(x1)))] = [1] x1 + [4] 237.91/60.05 >= [1] x1 + [4] 237.91/60.05 = [A(B(C(x1)))] 237.91/60.05 237.91/60.05 [B(b(x1))] = [1] x1 + [4] 237.91/60.05 > [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [B(C(a(x1)))] = [1] x1 + [4] 237.91/60.05 >= [1] x1 + [4] 237.91/60.05 = [a(C(B(x1)))] 237.91/60.05 237.91/60.05 [A(a(x1))] = [1] x1 + [4] 237.91/60.05 > [1] x1 + [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [A(c(b(x1)))] = [1] x1 + [4] 237.91/60.05 >= [1] x1 + [4] 237.91/60.05 = [b(c(A(x1)))] 237.91/60.05 237.91/60.05 237.91/60.05 We return to the main proof. 237.91/60.05 237.91/60.05 We are left with following problem, upon which TcT provides the 237.91/60.05 certificate YES(O(1),O(n^2)). 237.91/60.05 237.91/60.05 Strict Trs: 237.91/60.05 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.05 , b(a(C(x1))) -> C(a(b(x1))) 237.91/60.05 , c(A(B(x1))) -> B(A(c(x1))) 237.91/60.05 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.05 , B(C(a(x1))) -> a(C(B(x1))) 237.91/60.05 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.05 Weak Trs: 237.91/60.05 { a(A(x1)) -> x1 237.91/60.05 , b(B(x1)) -> x1 237.91/60.05 , c(C(x1)) -> x1 237.91/60.05 , C(c(x1)) -> x1 237.91/60.05 , B(b(x1)) -> x1 237.91/60.05 , A(a(x1)) -> x1 } 237.91/60.05 Obligation: 237.91/60.05 derivational complexity 237.91/60.05 Answer: 237.91/60.05 YES(O(1),O(n^2)) 237.91/60.05 237.91/60.05 We use the processor 'matrix interpretation of dimension 2' to 237.91/60.05 orient following rules strictly. 237.91/60.05 237.91/60.05 Trs: { A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.05 237.91/60.05 The induced complexity on above rules (modulo remaining rules) is 237.91/60.05 YES(?,O(n^2)) . These rules are moved into the corresponding weak 237.91/60.05 component(s). 237.91/60.05 237.91/60.05 Sub-proof: 237.91/60.05 ---------- 237.91/60.05 TcT has computed the following triangular matrix interpretation. 237.91/60.05 237.91/60.05 [a](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [b](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 237.91/60.05 [c](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [C](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [B](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [A](x1) = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 The order satisfies the following ordering constraints: 237.91/60.05 237.91/60.05 [a(b(c(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [c(b(a(x1)))] 237.91/60.05 237.91/60.05 [a(A(x1))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [b(a(C(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [C(a(b(x1)))] 237.91/60.05 237.91/60.05 [b(B(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(C(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(A(B(x1)))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [B(A(c(x1)))] 237.91/60.05 237.91/60.05 [C(c(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [C(B(A(x1)))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [A(B(C(x1)))] 237.91/60.05 237.91/60.05 [B(b(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [B(C(a(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [a(C(B(x1)))] 237.91/60.05 237.91/60.05 [A(a(x1))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [A(c(b(x1)))] = [1 1] x1 + [1] 237.91/60.05 [0 1] [1] 237.91/60.05 > [1 1] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [b(c(A(x1)))] 237.91/60.05 237.91/60.05 237.91/60.05 We return to the main proof. 237.91/60.05 237.91/60.05 We are left with following problem, upon which TcT provides the 237.91/60.05 certificate YES(O(1),O(n^2)). 237.91/60.05 237.91/60.05 Strict Trs: 237.91/60.05 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.05 , b(a(C(x1))) -> C(a(b(x1))) 237.91/60.05 , c(A(B(x1))) -> B(A(c(x1))) 237.91/60.05 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.05 , B(C(a(x1))) -> a(C(B(x1))) } 237.91/60.05 Weak Trs: 237.91/60.05 { a(A(x1)) -> x1 237.91/60.05 , b(B(x1)) -> x1 237.91/60.05 , c(C(x1)) -> x1 237.91/60.05 , C(c(x1)) -> x1 237.91/60.05 , B(b(x1)) -> x1 237.91/60.05 , A(a(x1)) -> x1 237.91/60.05 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.05 Obligation: 237.91/60.05 derivational complexity 237.91/60.05 Answer: 237.91/60.05 YES(O(1),O(n^2)) 237.91/60.05 237.91/60.05 We use the processor 'matrix interpretation of dimension 2' to 237.91/60.05 orient following rules strictly. 237.91/60.05 237.91/60.05 Trs: { a(b(c(x1))) -> c(b(a(x1))) } 237.91/60.05 237.91/60.05 The induced complexity on above rules (modulo remaining rules) is 237.91/60.05 YES(?,O(n^2)) . These rules are moved into the corresponding weak 237.91/60.05 component(s). 237.91/60.05 237.91/60.05 Sub-proof: 237.91/60.05 ---------- 237.91/60.05 TcT has computed the following triangular matrix interpretation. 237.91/60.05 237.91/60.05 [a](x1) = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [b](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [c](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 237.91/60.05 [C](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [B](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [A](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 The order satisfies the following ordering constraints: 237.91/60.05 237.91/60.05 [a(b(c(x1)))] = [1 1] x1 + [1] 237.91/60.05 [0 1] [1] 237.91/60.05 > [1 1] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [c(b(a(x1)))] 237.91/60.05 237.91/60.05 [a(A(x1))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [b(a(C(x1)))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [C(a(b(x1)))] 237.91/60.05 237.91/60.05 [b(B(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(C(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(A(B(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [B(A(c(x1)))] 237.91/60.05 237.91/60.05 [C(c(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [C(B(A(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [A(B(C(x1)))] 237.91/60.05 237.91/60.05 [B(b(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [B(C(a(x1)))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [a(C(B(x1)))] 237.91/60.05 237.91/60.05 [A(a(x1))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [A(c(b(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [b(c(A(x1)))] 237.91/60.05 237.91/60.05 237.91/60.05 We return to the main proof. 237.91/60.05 237.91/60.05 We are left with following problem, upon which TcT provides the 237.91/60.05 certificate YES(O(1),O(n^2)). 237.91/60.05 237.91/60.05 Strict Trs: 237.91/60.05 { b(a(C(x1))) -> C(a(b(x1))) 237.91/60.05 , c(A(B(x1))) -> B(A(c(x1))) 237.91/60.05 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.05 , B(C(a(x1))) -> a(C(B(x1))) } 237.91/60.05 Weak Trs: 237.91/60.05 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.05 , a(A(x1)) -> x1 237.91/60.05 , b(B(x1)) -> x1 237.91/60.05 , c(C(x1)) -> x1 237.91/60.05 , C(c(x1)) -> x1 237.91/60.05 , B(b(x1)) -> x1 237.91/60.05 , A(a(x1)) -> x1 237.91/60.05 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.05 Obligation: 237.91/60.05 derivational complexity 237.91/60.05 Answer: 237.91/60.05 YES(O(1),O(n^2)) 237.91/60.05 237.91/60.05 We use the processor 'matrix interpretation of dimension 2' to 237.91/60.05 orient following rules strictly. 237.91/60.05 237.91/60.05 Trs: { b(a(C(x1))) -> C(a(b(x1))) } 237.91/60.05 237.91/60.05 The induced complexity on above rules (modulo remaining rules) is 237.91/60.05 YES(?,O(n^2)) . These rules are moved into the corresponding weak 237.91/60.05 component(s). 237.91/60.05 237.91/60.05 Sub-proof: 237.91/60.05 ---------- 237.91/60.05 TcT has computed the following triangular matrix interpretation. 237.91/60.05 237.91/60.05 [a](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [b](x1) = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [c](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [C](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 237.91/60.05 [B](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [A](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 The order satisfies the following ordering constraints: 237.91/60.05 237.91/60.05 [a(b(c(x1)))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [c(b(a(x1)))] 237.91/60.05 237.91/60.05 [a(A(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [b(a(C(x1)))] = [1 1] x1 + [1] 237.91/60.05 [0 1] [1] 237.91/60.05 > [1 1] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [C(a(b(x1)))] 237.91/60.05 237.91/60.05 [b(B(x1))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(C(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(A(B(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [B(A(c(x1)))] 237.91/60.05 237.91/60.05 [C(c(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [C(B(A(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [A(B(C(x1)))] 237.91/60.05 237.91/60.05 [B(b(x1))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [B(C(a(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [a(C(B(x1)))] 237.91/60.05 237.91/60.05 [A(a(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [A(c(b(x1)))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [b(c(A(x1)))] 237.91/60.05 237.91/60.05 237.91/60.05 We return to the main proof. 237.91/60.05 237.91/60.05 We are left with following problem, upon which TcT provides the 237.91/60.05 certificate YES(O(1),O(n^2)). 237.91/60.05 237.91/60.05 Strict Trs: 237.91/60.05 { c(A(B(x1))) -> B(A(c(x1))) 237.91/60.05 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.05 , B(C(a(x1))) -> a(C(B(x1))) } 237.91/60.05 Weak Trs: 237.91/60.05 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.05 , a(A(x1)) -> x1 237.91/60.05 , b(a(C(x1))) -> C(a(b(x1))) 237.91/60.05 , b(B(x1)) -> x1 237.91/60.05 , c(C(x1)) -> x1 237.91/60.05 , C(c(x1)) -> x1 237.91/60.05 , B(b(x1)) -> x1 237.91/60.05 , A(a(x1)) -> x1 237.91/60.05 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.05 Obligation: 237.91/60.05 derivational complexity 237.91/60.05 Answer: 237.91/60.05 YES(O(1),O(n^2)) 237.91/60.05 237.91/60.05 We use the processor 'matrix interpretation of dimension 2' to 237.91/60.05 orient following rules strictly. 237.91/60.05 237.91/60.05 Trs: { C(B(A(x1))) -> A(B(C(x1))) } 237.91/60.05 237.91/60.05 The induced complexity on above rules (modulo remaining rules) is 237.91/60.05 YES(?,O(n^2)) . These rules are moved into the corresponding weak 237.91/60.05 component(s). 237.91/60.05 237.91/60.05 Sub-proof: 237.91/60.05 ---------- 237.91/60.05 TcT has computed the following triangular matrix interpretation. 237.91/60.05 237.91/60.05 [a](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [b](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [c](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [C](x1) = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [B](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 237.91/60.05 [A](x1) = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 237.91/60.05 The order satisfies the following ordering constraints: 237.91/60.05 237.91/60.05 [a(b(c(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [c(b(a(x1)))] 237.91/60.05 237.91/60.05 [a(A(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [b(a(C(x1)))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [C(a(b(x1)))] 237.91/60.05 237.91/60.05 [b(B(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(C(x1))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [c(A(B(x1)))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [B(A(c(x1)))] 237.91/60.05 237.91/60.05 [C(c(x1))] = [1 1] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.05 [C(B(A(x1)))] = [1 1] x1 + [1] 237.91/60.05 [0 1] [1] 237.91/60.05 > [1 1] x1 + [0] 237.91/60.05 [0 1] [1] 237.91/60.05 = [A(B(C(x1)))] 237.91/60.05 237.91/60.05 [B(b(x1))] = [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 >= [1 0] x1 + [0] 237.91/60.05 [0 1] [0] 237.91/60.05 = [x1] 237.91/60.05 237.91/60.06 [B(C(a(x1)))] = [1 1] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 1] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [a(C(B(x1)))] 237.91/60.06 237.91/60.06 [A(a(x1))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [1] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [A(c(b(x1)))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [1] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [1] 237.91/60.06 = [b(c(A(x1)))] 237.91/60.06 237.91/60.06 237.91/60.06 We return to the main proof. 237.91/60.06 237.91/60.06 We are left with following problem, upon which TcT provides the 237.91/60.06 certificate YES(O(1),O(n^2)). 237.91/60.06 237.91/60.06 Strict Trs: 237.91/60.06 { c(A(B(x1))) -> B(A(c(x1))) 237.91/60.06 , B(C(a(x1))) -> a(C(B(x1))) } 237.91/60.06 Weak Trs: 237.91/60.06 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.06 , a(A(x1)) -> x1 237.91/60.06 , b(a(C(x1))) -> C(a(b(x1))) 237.91/60.06 , b(B(x1)) -> x1 237.91/60.06 , c(C(x1)) -> x1 237.91/60.06 , C(c(x1)) -> x1 237.91/60.06 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.06 , B(b(x1)) -> x1 237.91/60.06 , A(a(x1)) -> x1 237.91/60.06 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.06 Obligation: 237.91/60.06 derivational complexity 237.91/60.06 Answer: 237.91/60.06 YES(O(1),O(n^2)) 237.91/60.06 237.91/60.06 We use the processor 'matrix interpretation of dimension 2' to 237.91/60.06 orient following rules strictly. 237.91/60.06 237.91/60.06 Trs: { B(C(a(x1))) -> a(C(B(x1))) } 237.91/60.06 237.91/60.06 The induced complexity on above rules (modulo remaining rules) is 237.91/60.06 YES(?,O(n^2)) . These rules are moved into the corresponding weak 237.91/60.06 component(s). 237.91/60.06 237.91/60.06 Sub-proof: 237.91/60.06 ---------- 237.91/60.06 TcT has computed the following triangular matrix interpretation. 237.91/60.06 237.91/60.06 [a](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 237.91/60.06 [b](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 237.91/60.06 [c](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 237.91/60.06 [C](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 237.91/60.06 [B](x1) = [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 237.91/60.06 [A](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 237.91/60.06 The order satisfies the following ordering constraints: 237.91/60.06 237.91/60.06 [a(b(c(x1)))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [4] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [4] 237.91/60.06 = [c(b(a(x1)))] 237.91/60.06 237.91/60.06 [a(A(x1))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [b(a(C(x1)))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [4] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [4] 237.91/60.06 = [C(a(b(x1)))] 237.91/60.06 237.91/60.06 [b(B(x1))] = [1 2] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [c(C(x1))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [c(A(B(x1)))] = [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [B(A(c(x1)))] 237.91/60.06 237.91/60.06 [C(c(x1))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [C(B(A(x1)))] = [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [A(B(C(x1)))] 237.91/60.06 237.91/60.06 [B(b(x1))] = [1 2] x1 + [4] 237.91/60.06 [0 1] [2] 237.91/60.06 > [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [B(C(a(x1)))] = [1 2] x1 + [4] 237.91/60.06 [0 1] [2] 237.91/60.06 > [1 2] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 = [a(C(B(x1)))] 237.91/60.06 237.91/60.06 [A(a(x1))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [A(c(b(x1)))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 = [b(c(A(x1)))] 237.91/60.06 237.91/60.06 237.91/60.06 We return to the main proof. 237.91/60.06 237.91/60.06 We are left with following problem, upon which TcT provides the 237.91/60.06 certificate YES(O(1),O(n^2)). 237.91/60.06 237.91/60.06 Strict Trs: { c(A(B(x1))) -> B(A(c(x1))) } 237.91/60.06 Weak Trs: 237.91/60.06 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.06 , a(A(x1)) -> x1 237.91/60.06 , b(a(C(x1))) -> C(a(b(x1))) 237.91/60.06 , b(B(x1)) -> x1 237.91/60.06 , c(C(x1)) -> x1 237.91/60.06 , C(c(x1)) -> x1 237.91/60.06 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.06 , B(b(x1)) -> x1 237.91/60.06 , B(C(a(x1))) -> a(C(B(x1))) 237.91/60.06 , A(a(x1)) -> x1 237.91/60.06 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.06 Obligation: 237.91/60.06 derivational complexity 237.91/60.06 Answer: 237.91/60.06 YES(O(1),O(n^2)) 237.91/60.06 237.91/60.06 We use the processor 'matrix interpretation of dimension 2' to 237.91/60.06 orient following rules strictly. 237.91/60.06 237.91/60.06 Trs: { c(A(B(x1))) -> B(A(c(x1))) } 237.91/60.06 237.91/60.06 The induced complexity on above rules (modulo remaining rules) is 237.91/60.06 YES(?,O(n^2)) . These rules are moved into the corresponding weak 237.91/60.06 component(s). 237.91/60.06 237.91/60.06 Sub-proof: 237.91/60.06 ---------- 237.91/60.06 TcT has computed the following triangular matrix interpretation. 237.91/60.06 237.91/60.06 [a](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 237.91/60.06 [b](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 237.91/60.06 [c](x1) = [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 237.91/60.06 [C](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 237.91/60.06 [B](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 237.91/60.06 [A](x1) = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 237.91/60.06 The order satisfies the following ordering constraints: 237.91/60.06 237.91/60.06 [a(b(c(x1)))] = [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [c(b(a(x1)))] 237.91/60.06 237.91/60.06 [a(A(x1))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [b(a(C(x1)))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [C(a(b(x1)))] 237.91/60.06 237.91/60.06 [b(B(x1))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [c(C(x1))] = [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [c(A(B(x1)))] = [1 2] x1 + [4] 237.91/60.06 [0 1] [2] 237.91/60.06 > [1 2] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 = [B(A(c(x1)))] 237.91/60.06 237.91/60.06 [C(c(x1))] = [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [C(B(A(x1)))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 = [A(B(C(x1)))] 237.91/60.06 237.91/60.06 [B(b(x1))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [B(C(a(x1)))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [2] 237.91/60.06 = [a(C(B(x1)))] 237.91/60.06 237.91/60.06 [A(a(x1))] = [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 0] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [x1] 237.91/60.06 237.91/60.06 [A(c(b(x1)))] = [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 >= [1 2] x1 + [0] 237.91/60.06 [0 1] [0] 237.91/60.06 = [b(c(A(x1)))] 237.91/60.06 237.91/60.06 237.91/60.06 We return to the main proof. 237.91/60.06 237.91/60.06 We are left with following problem, upon which TcT provides the 237.91/60.06 certificate YES(O(1),O(1)). 237.91/60.06 237.91/60.06 Weak Trs: 237.91/60.06 { a(b(c(x1))) -> c(b(a(x1))) 237.91/60.06 , a(A(x1)) -> x1 237.91/60.06 , b(a(C(x1))) -> C(a(b(x1))) 237.91/60.06 , b(B(x1)) -> x1 237.91/60.06 , c(C(x1)) -> x1 237.91/60.06 , c(A(B(x1))) -> B(A(c(x1))) 237.91/60.06 , C(c(x1)) -> x1 237.91/60.06 , C(B(A(x1))) -> A(B(C(x1))) 237.91/60.06 , B(b(x1)) -> x1 237.91/60.06 , B(C(a(x1))) -> a(C(B(x1))) 237.91/60.06 , A(a(x1)) -> x1 237.91/60.06 , A(c(b(x1))) -> b(c(A(x1))) } 237.91/60.06 Obligation: 237.91/60.06 derivational complexity 237.91/60.06 Answer: 237.91/60.06 YES(O(1),O(1)) 237.91/60.06 237.91/60.06 Empty rules are trivially bounded 237.91/60.06 237.91/60.06 Hurray, we answered YES(O(1),O(n^2)) 237.91/60.06 EOF