Problem Transformed CSR 04 Ex15 Luc06 Z

interpretations

Execution Time (secs)
-
Answer
YES(?,O(n^3))
InputTransformed CSR 04 Ex15 Luc06 Z
YES(?,O(n^3))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict Trs:
  { f(X) -> n__f(X)
  , f(n__f(n__a())) -> f(n__g(f(n__a())))
  , a() -> n__a()
  , g(X) -> n__g(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

The following argument positions are usable:
  Uargs(f) = {1}, Uargs(n__f) = {}, Uargs(n__g) = {1}, Uargs(g) = {},
  Uargs(activate) = {}

TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).

                   [3 1 0]      [1]
         [f](x1) = [0 1 1] x1 + [3]
                   [0 1 1]      [1]
                                   
                   [1 1 0]      [0]
      [n__f](x1) = [0 0 1] x1 + [3]
                   [0 1 0]      [0]
                                   
                   [0]
          [n__a] = [2]
                   [3]
                      
                   [1 0 0]      [0]
      [n__g](x1) = [0 0 0] x1 + [2]
                   [0 0 0]      [0]
                                   
                   [1]
             [a] = [2]
                   [3]
                      
                   [1 0 0]      [3]
         [g](x1) = [0 0 0] x1 + [2]
                   [0 0 0]      [0]
                                   
                   [3 3 1]      [2]
  [activate](x1) = [0 2 1] x1 + [2]
                   [0 1 1]      [0]

This order satisfies following ordering constraints

               [f(X)] = [3 1 0]     [1]     
                        [0 1 1] X + [3]     
                        [0 1 1]     [1]     
                      > [1 1 0]     [0]     
                        [0 0 1] X + [3]     
                        [0 1 0]     [0]     
                      = [n__f(X)]           
                                            
    [f(n__f(n__a()))] = [13]                
                        [11]                
                        [9]                 
                      > [12]                
                        [5]                 
                        [3]                 
                      = [f(n__g(f(n__a())))]
                                            
                [a()] = [1]                 
                        [2]                 
                        [3]                 
                      > [0]                 
                        [2]                 
                        [3]                 
                      = [n__a()]            
                                            
               [g(X)] = [1 0 0]     [3]     
                        [0 0 0] X + [2]     
                        [0 0 0]     [0]     
                      > [1 0 0]     [0]     
                        [0 0 0] X + [2]     
                        [0 0 0]     [0]     
                      = [n__g(X)]           
                                            
        [activate(X)] = [3 3 1]     [2]     
                        [0 2 1] X + [2]     
                        [0 1 1]     [0]     
                      > [1 0 0]     [0]     
                        [0 1 0] X + [0]     
                        [0 0 1]     [0]     
                      = [X]                 
                                            
  [activate(n__f(X))] = [3 4 3]     [11]    
                        [0 1 2] X + [8]     
                        [0 1 1]     [3]     
                      > [3 1 0]     [1]     
                        [0 1 1] X + [3]     
                        [0 1 1]     [1]     
                      = [f(X)]              
                                            
   [activate(n__a())] = [11]                
                        [9]                 
                        [5]                 
                      > [1]                 
                        [2]                 
                        [3]                 
                      = [a()]               
                                            
  [activate(n__g(X))] = [3 0 0]     [8]     
                        [0 0 0] X + [6]     
                        [0 0 0]     [2]     
                      > [1 0 0]     [3]     
                        [0 0 0] X + [2]     
                        [0 0 0]     [0]     
                      = [g(X)]              
                                            

Hurray, we answered YES(?,O(n^3))

lmpo

Execution Time (secs)
-
Answer
MAYBE
InputTransformed CSR 04 Ex15 Luc06 Z
MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { f(n__f(n__a())) -> f(n__g(f(n__a())))
  , f(X) -> n__f(X)
  , a() -> n__a()
  , g(X) -> n__g(X)
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(X)
  , activate(X) -> X }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..

mpo

Execution Time (secs)
-
Answer
MAYBE
InputTransformed CSR 04 Ex15 Luc06 Z
MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { f(n__f(n__a())) -> f(n__g(f(n__a())))
  , f(X) -> n__f(X)
  , a() -> n__a()
  , g(X) -> n__g(X)
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(X)
  , activate(X) -> X }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..

popstar

Execution Time (secs)
0.122
Answer
MAYBE
InputTransformed CSR 04 Ex15 Luc06 Z
MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { f(n__f(n__a())) -> f(n__g(f(n__a())))
  , f(X) -> n__f(X)
  , a() -> n__a()
  , g(X) -> n__g(X)
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(X)
  , activate(X) -> X }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..

popstar-ps

Execution Time (secs)
0.205
Answer
MAYBE
InputTransformed CSR 04 Ex15 Luc06 Z
MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { f(n__f(n__a())) -> f(n__g(f(n__a())))
  , f(X) -> n__f(X)
  , a() -> n__a()
  , g(X) -> n__g(X)
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(X)
  , activate(X) -> X }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..