YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { fst(X1, X2) -> n__fst(X1, X2) , fst(0(), Z) -> nil() , fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) , activate(X) -> X , activate(n__fst(X1, X2)) -> fst(X1, X2) , activate(n__from(X)) -> from(X) , activate(n__add(X1, X2)) -> add(X1, X2) , activate(n__len(X)) -> len(X) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , add(X1, X2) -> n__add(X1, X2) , add(0(), X) -> X , add(s(X), Y) -> s(n__add(activate(X), Y)) , len(X) -> n__len(X) , len(nil()) -> 0() , len(cons(X, Z)) -> s(n__len(activate(Z))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { fst(X1, X2) -> n__fst(X1, X2) , fst(0(), Z) -> nil() , fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) , activate(X) -> X , from(X) -> n__from(X) , add(X1, X2) -> n__add(X1, X2) , add(0(), X) -> X , len(X) -> n__len(X) , len(nil()) -> 0() , len(cons(X, Z)) -> s(n__len(activate(Z))) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [fst](x1, x2) = [3] x1 + [3] x2 + [1] [0] = [0] [nil] = [0] [s](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [1] [n__fst](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [3] x1 + [1] [from](x1) = [3] x1 + [1] [n__from](x1) = [1] x1 + [0] [add](x1, x2) = [3] x1 + [3] x2 + [1] [n__add](x1, x2) = [1] x1 + [1] x2 + [0] [len](x1) = [3] x1 + [1] [n__len](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [fst(X1, X2)] = [3] X1 + [3] X2 + [1] > [1] X1 + [1] X2 + [0] = [n__fst(X1, X2)] [fst(0(), Z)] = [3] Z + [1] > [0] = [nil()] [fst(s(X), cons(Y, Z))] = [3] Z + [3] X + [3] Y + [4] > [3] Z + [3] X + [1] Y + [3] = [cons(Y, n__fst(activate(X), activate(Z)))] [activate(X)] = [3] X + [1] > [1] X + [0] = [X] [activate(n__fst(X1, X2))] = [3] X1 + [3] X2 + [1] >= [3] X1 + [3] X2 + [1] = [fst(X1, X2)] [activate(n__from(X))] = [3] X + [1] >= [3] X + [1] = [from(X)] [activate(n__add(X1, X2))] = [3] X1 + [3] X2 + [1] >= [3] X1 + [3] X2 + [1] = [add(X1, X2)] [activate(n__len(X))] = [3] X + [1] >= [3] X + [1] = [len(X)] [from(X)] = [3] X + [1] >= [2] X + [1] = [cons(X, n__from(s(X)))] [from(X)] = [3] X + [1] > [1] X + [0] = [n__from(X)] [add(X1, X2)] = [3] X1 + [3] X2 + [1] > [1] X1 + [1] X2 + [0] = [n__add(X1, X2)] [add(0(), X)] = [3] X + [1] > [1] X + [0] = [X] [add(s(X), Y)] = [3] X + [3] Y + [1] >= [3] X + [1] Y + [1] = [s(n__add(activate(X), Y))] [len(X)] = [3] X + [1] > [1] X + [0] = [n__len(X)] [len(nil())] = [1] > [0] = [0()] [len(cons(X, Z))] = [3] Z + [3] X + [4] > [3] Z + [1] = [s(n__len(activate(Z)))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { activate(n__fst(X1, X2)) -> fst(X1, X2) , activate(n__from(X)) -> from(X) , activate(n__add(X1, X2)) -> add(X1, X2) , activate(n__len(X)) -> len(X) , from(X) -> cons(X, n__from(s(X))) , add(s(X), Y) -> s(n__add(activate(X), Y)) } Weak Trs: { fst(X1, X2) -> n__fst(X1, X2) , fst(0(), Z) -> nil() , fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) , activate(X) -> X , from(X) -> n__from(X) , add(X1, X2) -> n__add(X1, X2) , add(0(), X) -> X , len(X) -> n__len(X) , len(nil()) -> 0() , len(cons(X, Z)) -> s(n__len(activate(Z))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { activate(n__add(X1, X2)) -> add(X1, X2) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [fst](x1, x2) = [2] x1 + [2] x2 + [0] [0] = [0] [nil] = [0] [s](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [n__fst](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [2] x1 + [0] [from](x1) = [2] x1 + [0] [n__from](x1) = [1] x1 + [0] [add](x1, x2) = [2] x1 + [1] x2 + [2] [n__add](x1, x2) = [1] x1 + [1] x2 + [2] [len](x1) = [2] x1 + [0] [n__len](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [fst(X1, X2)] = [2] X1 + [2] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__fst(X1, X2)] [fst(0(), Z)] = [2] Z + [0] >= [0] = [nil()] [fst(s(X), cons(Y, Z))] = [2] Z + [2] X + [2] Y + [0] >= [2] Z + [2] X + [1] Y + [0] = [cons(Y, n__fst(activate(X), activate(Z)))] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__fst(X1, X2))] = [2] X1 + [2] X2 + [0] >= [2] X1 + [2] X2 + [0] = [fst(X1, X2)] [activate(n__from(X))] = [2] X + [0] >= [2] X + [0] = [from(X)] [activate(n__add(X1, X2))] = [2] X1 + [2] X2 + [4] > [2] X1 + [1] X2 + [2] = [add(X1, X2)] [activate(n__len(X))] = [2] X + [0] >= [2] X + [0] = [len(X)] [from(X)] = [2] X + [0] >= [2] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [2] X + [0] >= [1] X + [0] = [n__from(X)] [add(X1, X2)] = [2] X1 + [1] X2 + [2] >= [1] X1 + [1] X2 + [2] = [n__add(X1, X2)] [add(0(), X)] = [1] X + [2] > [1] X + [0] = [X] [add(s(X), Y)] = [2] X + [1] Y + [2] >= [2] X + [1] Y + [2] = [s(n__add(activate(X), Y))] [len(X)] = [2] X + [0] >= [1] X + [0] = [n__len(X)] [len(nil())] = [0] >= [0] = [0()] [len(cons(X, Z))] = [2] Z + [2] X + [0] >= [2] Z + [0] = [s(n__len(activate(Z)))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { activate(n__fst(X1, X2)) -> fst(X1, X2) , activate(n__from(X)) -> from(X) , activate(n__len(X)) -> len(X) , from(X) -> cons(X, n__from(s(X))) , add(s(X), Y) -> s(n__add(activate(X), Y)) } Weak Trs: { fst(X1, X2) -> n__fst(X1, X2) , fst(0(), Z) -> nil() , fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) , activate(X) -> X , activate(n__add(X1, X2)) -> add(X1, X2) , from(X) -> n__from(X) , add(X1, X2) -> n__add(X1, X2) , add(0(), X) -> X , len(X) -> n__len(X) , len(nil()) -> 0() , len(cons(X, Z)) -> s(n__len(activate(Z))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { add(s(X), Y) -> s(n__add(activate(X), Y)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [fst](x1, x2) = [2] x1 + [2] x2 + [0] [0] = [0] [nil] = [0] [s](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [n__fst](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [2] x1 + [0] [from](x1) = [2] x1 + [0] [n__from](x1) = [1] x1 + [0] [add](x1, x2) = [2] x1 + [1] x2 + [3] [n__add](x1, x2) = [1] x1 + [1] x2 + [2] [len](x1) = [2] x1 + [0] [n__len](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [fst(X1, X2)] = [2] X1 + [2] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__fst(X1, X2)] [fst(0(), Z)] = [2] Z + [0] >= [0] = [nil()] [fst(s(X), cons(Y, Z))] = [2] Z + [2] X + [2] Y + [0] >= [2] Z + [2] X + [1] Y + [0] = [cons(Y, n__fst(activate(X), activate(Z)))] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__fst(X1, X2))] = [2] X1 + [2] X2 + [0] >= [2] X1 + [2] X2 + [0] = [fst(X1, X2)] [activate(n__from(X))] = [2] X + [0] >= [2] X + [0] = [from(X)] [activate(n__add(X1, X2))] = [2] X1 + [2] X2 + [4] > [2] X1 + [1] X2 + [3] = [add(X1, X2)] [activate(n__len(X))] = [2] X + [0] >= [2] X + [0] = [len(X)] [from(X)] = [2] X + [0] >= [2] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [2] X + [0] >= [1] X + [0] = [n__from(X)] [add(X1, X2)] = [2] X1 + [1] X2 + [3] > [1] X1 + [1] X2 + [2] = [n__add(X1, X2)] [add(0(), X)] = [1] X + [3] > [1] X + [0] = [X] [add(s(X), Y)] = [2] X + [1] Y + [3] > [2] X + [1] Y + [2] = [s(n__add(activate(X), Y))] [len(X)] = [2] X + [0] >= [1] X + [0] = [n__len(X)] [len(nil())] = [0] >= [0] = [0()] [len(cons(X, Z))] = [2] Z + [2] X + [0] >= [2] Z + [0] = [s(n__len(activate(Z)))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { activate(n__fst(X1, X2)) -> fst(X1, X2) , activate(n__from(X)) -> from(X) , activate(n__len(X)) -> len(X) , from(X) -> cons(X, n__from(s(X))) } Weak Trs: { fst(X1, X2) -> n__fst(X1, X2) , fst(0(), Z) -> nil() , fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) , activate(X) -> X , activate(n__add(X1, X2)) -> add(X1, X2) , from(X) -> n__from(X) , add(X1, X2) -> n__add(X1, X2) , add(0(), X) -> X , add(s(X), Y) -> s(n__add(activate(X), Y)) , len(X) -> n__len(X) , len(nil()) -> 0() , len(cons(X, Z)) -> s(n__len(activate(Z))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { activate(n__len(X)) -> len(X) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [fst](x1, x2) = [3] x1 + [3] x2 + [1] [0] = [0] [nil] = [0] [s](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [1] [n__fst](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [3] x1 + [1] [from](x1) = [3] x1 + [1] [n__from](x1) = [1] x1 + [0] [add](x1, x2) = [3] x1 + [3] x2 + [1] [n__add](x1, x2) = [1] x1 + [1] x2 + [0] [len](x1) = [3] x1 + [0] [n__len](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [fst(X1, X2)] = [3] X1 + [3] X2 + [1] > [1] X1 + [1] X2 + [0] = [n__fst(X1, X2)] [fst(0(), Z)] = [3] Z + [1] > [0] = [nil()] [fst(s(X), cons(Y, Z))] = [3] Z + [3] X + [3] Y + [4] > [3] Z + [3] X + [1] Y + [3] = [cons(Y, n__fst(activate(X), activate(Z)))] [activate(X)] = [3] X + [1] > [1] X + [0] = [X] [activate(n__fst(X1, X2))] = [3] X1 + [3] X2 + [1] >= [3] X1 + [3] X2 + [1] = [fst(X1, X2)] [activate(n__from(X))] = [3] X + [1] >= [3] X + [1] = [from(X)] [activate(n__add(X1, X2))] = [3] X1 + [3] X2 + [1] >= [3] X1 + [3] X2 + [1] = [add(X1, X2)] [activate(n__len(X))] = [3] X + [1] > [3] X + [0] = [len(X)] [from(X)] = [3] X + [1] >= [2] X + [1] = [cons(X, n__from(s(X)))] [from(X)] = [3] X + [1] > [1] X + [0] = [n__from(X)] [add(X1, X2)] = [3] X1 + [3] X2 + [1] > [1] X1 + [1] X2 + [0] = [n__add(X1, X2)] [add(0(), X)] = [3] X + [1] > [1] X + [0] = [X] [add(s(X), Y)] = [3] X + [3] Y + [1] >= [3] X + [1] Y + [1] = [s(n__add(activate(X), Y))] [len(X)] = [3] X + [0] >= [1] X + [0] = [n__len(X)] [len(nil())] = [0] >= [0] = [0()] [len(cons(X, Z))] = [3] Z + [3] X + [3] > [3] Z + [1] = [s(n__len(activate(Z)))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { activate(n__fst(X1, X2)) -> fst(X1, X2) , activate(n__from(X)) -> from(X) , from(X) -> cons(X, n__from(s(X))) } Weak Trs: { fst(X1, X2) -> n__fst(X1, X2) , fst(0(), Z) -> nil() , fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) , activate(X) -> X , activate(n__add(X1, X2)) -> add(X1, X2) , activate(n__len(X)) -> len(X) , from(X) -> n__from(X) , add(X1, X2) -> n__add(X1, X2) , add(0(), X) -> X , add(s(X), Y) -> s(n__add(activate(X), Y)) , len(X) -> n__len(X) , len(nil()) -> 0() , len(cons(X, Z)) -> s(n__len(activate(Z))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { activate(n__fst(X1, X2)) -> fst(X1, X2) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [fst](x1, x2) = [2] x1 + [2] x2 + [0] [0] = [0] [nil] = [0] [s](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [2] [n__fst](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [2] x1 + [1] [from](x1) = [2] x1 + [3] [n__from](x1) = [1] x1 + [1] [add](x1, x2) = [2] x1 + [2] x2 + [1] [n__add](x1, x2) = [1] x1 + [1] x2 + [0] [len](x1) = [2] x1 + [0] [n__len](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [fst(X1, X2)] = [2] X1 + [2] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__fst(X1, X2)] [fst(0(), Z)] = [2] Z + [0] >= [0] = [nil()] [fst(s(X), cons(Y, Z))] = [2] Z + [2] X + [2] Y + [4] >= [2] Z + [2] X + [1] Y + [4] = [cons(Y, n__fst(activate(X), activate(Z)))] [activate(X)] = [2] X + [1] > [1] X + [0] = [X] [activate(n__fst(X1, X2))] = [2] X1 + [2] X2 + [1] > [2] X1 + [2] X2 + [0] = [fst(X1, X2)] [activate(n__from(X))] = [2] X + [3] >= [2] X + [3] = [from(X)] [activate(n__add(X1, X2))] = [2] X1 + [2] X2 + [1] >= [2] X1 + [2] X2 + [1] = [add(X1, X2)] [activate(n__len(X))] = [2] X + [1] > [2] X + [0] = [len(X)] [from(X)] = [2] X + [3] >= [2] X + [3] = [cons(X, n__from(s(X)))] [from(X)] = [2] X + [3] > [1] X + [1] = [n__from(X)] [add(X1, X2)] = [2] X1 + [2] X2 + [1] > [1] X1 + [1] X2 + [0] = [n__add(X1, X2)] [add(0(), X)] = [2] X + [1] > [1] X + [0] = [X] [add(s(X), Y)] = [2] X + [2] Y + [1] >= [2] X + [1] Y + [1] = [s(n__add(activate(X), Y))] [len(X)] = [2] X + [0] >= [1] X + [0] = [n__len(X)] [len(nil())] = [0] >= [0] = [0()] [len(cons(X, Z))] = [2] Z + [2] X + [4] > [2] Z + [1] = [s(n__len(activate(Z)))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { activate(n__from(X)) -> from(X) , from(X) -> cons(X, n__from(s(X))) } Weak Trs: { fst(X1, X2) -> n__fst(X1, X2) , fst(0(), Z) -> nil() , fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) , activate(X) -> X , activate(n__fst(X1, X2)) -> fst(X1, X2) , activate(n__add(X1, X2)) -> add(X1, X2) , activate(n__len(X)) -> len(X) , from(X) -> n__from(X) , add(X1, X2) -> n__add(X1, X2) , add(0(), X) -> X , add(s(X), Y) -> s(n__add(activate(X), Y)) , len(X) -> n__len(X) , len(nil()) -> 0() , len(cons(X, Z)) -> s(n__len(activate(Z))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { activate(n__from(X)) -> from(X) , from(X) -> cons(X, n__from(s(X))) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [fst](x1, x2) = [2] x1 + [2] x2 + [0] [0] = [0] [nil] = [0] [s](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [n__fst](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [2] x1 + [0] [from](x1) = [2] x1 + [3] [n__from](x1) = [1] x1 + [2] [add](x1, x2) = [2] x1 + [1] x2 + [0] [n__add](x1, x2) = [1] x1 + [1] x2 + [0] [len](x1) = [2] x1 + [0] [n__len](x1) = [1] x1 + [0] This order satisfies the following ordering constraints: [fst(X1, X2)] = [2] X1 + [2] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__fst(X1, X2)] [fst(0(), Z)] = [2] Z + [0] >= [0] = [nil()] [fst(s(X), cons(Y, Z))] = [2] Z + [2] X + [2] Y + [0] >= [2] Z + [2] X + [1] Y + [0] = [cons(Y, n__fst(activate(X), activate(Z)))] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__fst(X1, X2))] = [2] X1 + [2] X2 + [0] >= [2] X1 + [2] X2 + [0] = [fst(X1, X2)] [activate(n__from(X))] = [2] X + [4] > [2] X + [3] = [from(X)] [activate(n__add(X1, X2))] = [2] X1 + [2] X2 + [0] >= [2] X1 + [1] X2 + [0] = [add(X1, X2)] [activate(n__len(X))] = [2] X + [0] >= [2] X + [0] = [len(X)] [from(X)] = [2] X + [3] > [2] X + [2] = [cons(X, n__from(s(X)))] [from(X)] = [2] X + [3] > [1] X + [2] = [n__from(X)] [add(X1, X2)] = [2] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__add(X1, X2)] [add(0(), X)] = [1] X + [0] >= [1] X + [0] = [X] [add(s(X), Y)] = [2] X + [1] Y + [0] >= [2] X + [1] Y + [0] = [s(n__add(activate(X), Y))] [len(X)] = [2] X + [0] >= [1] X + [0] = [n__len(X)] [len(nil())] = [0] >= [0] = [0()] [len(cons(X, Z))] = [2] Z + [2] X + [0] >= [2] Z + [0] = [s(n__len(activate(Z)))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { fst(X1, X2) -> n__fst(X1, X2) , fst(0(), Z) -> nil() , fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) , activate(X) -> X , activate(n__fst(X1, X2)) -> fst(X1, X2) , activate(n__from(X)) -> from(X) , activate(n__add(X1, X2)) -> add(X1, X2) , activate(n__len(X)) -> len(X) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , add(X1, X2) -> n__add(X1, X2) , add(0(), X) -> X , add(s(X), Y) -> s(n__add(activate(X), Y)) , len(X) -> n__len(X) , len(nil()) -> 0() , len(cons(X, Z)) -> s(n__len(activate(Z))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))