YES(O(1),O(n^2))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { :(z, +(x, f(y))) -> :(g(z, y), +(x, a()))
  , :(:(x, y), z) -> :(x, :(y, z))
  , :(+(x, y), z) -> +(:(x, z), :(y, z)) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'polynomial interpretation' to orient
following rules strictly.

Trs: { :(+(x, y), z) -> +(:(x, z), :(y, z)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The following argument positions are considered usable:
  
    Uargs(:) = {2}, Uargs(+) = {1, 2}
  
  TcT has computed the following constructor-restricted polynomial
  interpretation.
  
  [:](x1, x2) = 2*x1 + 2*x1*x2 + x2
                                   
  [+](x1, x2) = 1 + x1 + x2        
                                   
      [f](x1) = x1                 
                                   
  [g](x1, x2) = 0                  
                                   
        [a]() = 0                  
                                   
  
  The following symbols are considered usable
  
    {:}
  
  This order satisfies the following ordering constraints.
  
    [:(z, +(x, f(y)))] =  4*z + 2*z*x + 2*z*y + 1 + x + y                
                       >= 1 + x                                          
                       =  [:(g(z, y), +(x, a()))]                        
                                                                         
       [:(:(x, y), z)] =  4*x + 4*x*y + 2*y + 4*x*z + 4*x*y*z + 2*y*z + z
                       >= 2*x + 4*x*y + 4*x*y*z + 2*x*z + 2*y + 2*y*z + z
                       =  [:(x, :(y, z))]                                
                                                                         
       [:(+(x, y), z)] =  2 + 2*x + 2*y + 3*z + 2*x*z + 2*y*z            
                       >  1 + 2*x + 2*x*z + 2*z + 2*y + 2*y*z            
                       =  [+(:(x, z), :(y, z))]                          
                                                                         

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { :(z, +(x, f(y))) -> :(g(z, y), +(x, a()))
  , :(:(x, y), z) -> :(x, :(y, z)) }
Weak Trs: { :(+(x, y), z) -> +(:(x, z), :(y, z)) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'polynomial interpretation' to orient
following rules strictly.

Trs: { :(:(x, y), z) -> :(x, :(y, z)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The following argument positions are considered usable:
  
    Uargs(:) = {2}, Uargs(+) = {1, 2}
  
  TcT has computed the following constructor-restricted polynomial
  interpretation.
  
  [:](x1, x2) = 1 + x1 + x1*x2 + x1^2 + x2
                                          
  [+](x1, x2) = 2 + x1 + x2               
                                          
      [f](x1) = 2 + x1                    
                                          
  [g](x1, x2) = 0                         
                                          
        [a]() = 2                         
                                          
  
  The following symbols are considered usable
  
    {:}
  
  This order satisfies the following ordering constraints.
  
    [:(z, +(x, f(y)))] =  5 + 5*z + z*x + z*y + z^2 + x + y                                                                                                              
                       >= 5 + x                                                                                                                                          
                       =  [:(g(z, y), +(x, a()))]                                                                                                                        
                                                                                                                                                                         
       [:(:(x, y), z)] =  3 + 3*x + 4*x*y + 4*x^2 + 3*y + 2*z + x*z + x*y*z + x^2*z + y*z + 3*x^2*y + 2*x^3 + x^2*y^2 + 2*x^3*y + x*y^2 + x^4 + y*x + y^2*x + y*x^2 + y^2
                       >  2 + 2*x + x*y + x*y*z + x*y^2 + x*z + x^2 + y + y*z + y^2 + z                                                                                  
                       =  [:(x, :(y, z))]                                                                                                                                
                                                                                                                                                                         
       [:(+(x, y), z)] =  7 + 5*x + 5*y + 3*z + x*z + y*z + x^2 + x*y + y*x + y^2                                                                                        
                       >  4 + x + x*z + x^2 + 2*z + y + y*z + y^2                                                                                                        
                       =  [+(:(x, z), :(y, z))]                                                                                                                          
                                                                                                                                                                         

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs: { :(z, +(x, f(y))) -> :(g(z, y), +(x, a())) }
Weak Trs:
  { :(:(x, y), z) -> :(x, :(y, z))
  , :(+(x, y), z) -> +(:(x, z), :(y, z)) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'polynomial interpretation' to orient
following rules strictly.

Trs: { :(z, +(x, f(y))) -> :(g(z, y), +(x, a())) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The following argument positions are considered usable:
  
    Uargs(:) = {2}, Uargs(+) = {1, 2}
  
  TcT has computed the following constructor-restricted polynomial
  interpretation.
  
  [:](x1, x2) = x1 + x1*x2 + x2
                               
  [+](x1, x2) = 1 + x1 + x2    
                               
      [f](x1) = 1 + x1         
                               
  [g](x1, x2) = 0              
                               
        [a]() = 0              
                               
  
  The following symbols are considered usable
  
    {:}
  
  This order satisfies the following ordering constraints.
  
    [:(z, +(x, f(y)))] =  3*z + z*x + z*y + 2 + x + y        
                       >  1 + x                              
                       =  [:(g(z, y), +(x, a()))]            
                                                             
       [:(:(x, y), z)] =  x + x*y + y + x*z + x*y*z + y*z + z
                       >= x + x*y + x*y*z + x*z + y + y*z + z
                       =  [:(x, :(y, z))]                    
                                                             
       [:(+(x, y), z)] =  1 + x + y + 2*z + x*z + y*z        
                       >= 1 + x + x*z + 2*z + y + y*z        
                       =  [+(:(x, z), :(y, 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:
  { :(z, +(x, f(y))) -> :(g(z, y), +(x, a()))
  , :(:(x, y), z) -> :(x, :(y, z))
  , :(+(x, y), z) -> +(:(x, z), :(y, z)) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

Hurray, we answered YES(O(1),O(n^2))