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:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(X))))
  , a__c(X) -> d(X)
  , a__c(X) -> c(X)
  , a__h(X) -> a__c(d(X))
  , a__h(X) -> h(X)
  , mark(f(X)) -> a__f(mark(X))
  , mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(X)
  , mark(h(X)) -> a__h(mark(X)) }
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:
  { mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(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).
  
    [a__f](x1) = [1] x1 + [2]
                             
       [f](x1) = [1] x1 + [2]
                             
    [a__c](x1) = [1] x1 + [0]
                             
       [g](x1) = [1] x1 + [0]
                             
       [d](x1) = [1] x1 + [0]
                             
    [a__h](x1) = [1] x1 + [0]
                             
    [mark](x1) = [1] x1 + [2]
                             
       [c](x1) = [1] x1 + [0]
                             
       [h](x1) = [1] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X)] =  [1] X + [2]       
                 >= [1] X + [2]       
                 =  [f(X)]            
                                      
    [a__f(f(X))] =  [1] X + [4]       
                 >= [1] X + [4]       
                 =  [a__c(f(g(f(X))))]
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [d(X)]            
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [c(X)]            
                                      
       [a__h(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [a__c(d(X))]      
                                      
       [a__h(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [h(X)]            
                                      
    [mark(f(X))] =  [1] X + [4]       
                 >= [1] X + [4]       
                 =  [a__f(mark(X))]   
                                      
    [mark(g(X))] =  [1] X + [2]       
                 >  [1] X + [0]       
                 =  [g(X)]            
                                      
    [mark(d(X))] =  [1] X + [2]       
                 >  [1] X + [0]       
                 =  [d(X)]            
                                      
    [mark(c(X))] =  [1] X + [2]       
                 >  [1] X + [0]       
                 =  [a__c(X)]         
                                      
    [mark(h(X))] =  [1] X + [2]       
                 >= [1] X + [2]       
                 =  [a__h(mark(X))]   
                                      

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:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(X))))
  , a__c(X) -> d(X)
  , a__c(X) -> c(X)
  , a__h(X) -> a__c(d(X))
  , a__h(X) -> h(X)
  , mark(f(X)) -> a__f(mark(X))
  , mark(h(X)) -> a__h(mark(X)) }
Weak Trs:
  { mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(X) }
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:
  { a__h(X) -> a__c(d(X))
  , mark(h(X)) -> a__h(mark(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).
  
    [a__f](x1) = [1] x1 + [0]
                             
       [f](x1) = [1] x1 + [0]
                             
    [a__c](x1) = [1] x1 + [0]
                             
       [g](x1) = [1] x1 + [0]
                             
       [d](x1) = [1] x1 + [0]
                             
    [a__h](x1) = [1] x1 + [2]
                             
    [mark](x1) = [2] x1 + [0]
                             
       [c](x1) = [1] x1 + [0]
                             
       [h](x1) = [1] x1 + [2]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [f(X)]            
                                      
    [a__f(f(X))] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [a__c(f(g(f(X))))]
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [d(X)]            
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [c(X)]            
                                      
       [a__h(X)] =  [1] X + [2]       
                 >  [1] X + [0]       
                 =  [a__c(d(X))]      
                                      
       [a__h(X)] =  [1] X + [2]       
                 >= [1] X + [2]       
                 =  [h(X)]            
                                      
    [mark(f(X))] =  [2] X + [0]       
                 >= [2] X + [0]       
                 =  [a__f(mark(X))]   
                                      
    [mark(g(X))] =  [2] X + [0]       
                 >= [1] X + [0]       
                 =  [g(X)]            
                                      
    [mark(d(X))] =  [2] X + [0]       
                 >= [1] X + [0]       
                 =  [d(X)]            
                                      
    [mark(c(X))] =  [2] X + [0]       
                 >= [1] X + [0]       
                 =  [a__c(X)]         
                                      
    [mark(h(X))] =  [2] X + [4]       
                 >  [2] X + [2]       
                 =  [a__h(mark(X))]   
                                      

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:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(X))))
  , a__c(X) -> d(X)
  , a__c(X) -> c(X)
  , a__h(X) -> h(X)
  , mark(f(X)) -> a__f(mark(X)) }
Weak Trs:
  { a__h(X) -> a__c(d(X))
  , mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(X)
  , mark(h(X)) -> a__h(mark(X)) }
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: { a__h(X) -> h(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).
  
    [a__f](x1) = [1] x1 + [0]
                             
       [f](x1) = [1] x1 + [0]
                             
    [a__c](x1) = [1] x1 + [0]
                             
       [g](x1) = [1] x1 + [0]
                             
       [d](x1) = [1] x1 + [0]
                             
    [a__h](x1) = [1] x1 + [3]
                             
    [mark](x1) = [2] x1 + [0]
                             
       [c](x1) = [1] x1 + [0]
                             
       [h](x1) = [1] x1 + [2]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [f(X)]            
                                      
    [a__f(f(X))] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [a__c(f(g(f(X))))]
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [d(X)]            
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [c(X)]            
                                      
       [a__h(X)] =  [1] X + [3]       
                 >  [1] X + [0]       
                 =  [a__c(d(X))]      
                                      
       [a__h(X)] =  [1] X + [3]       
                 >  [1] X + [2]       
                 =  [h(X)]            
                                      
    [mark(f(X))] =  [2] X + [0]       
                 >= [2] X + [0]       
                 =  [a__f(mark(X))]   
                                      
    [mark(g(X))] =  [2] X + [0]       
                 >= [1] X + [0]       
                 =  [g(X)]            
                                      
    [mark(d(X))] =  [2] X + [0]       
                 >= [1] X + [0]       
                 =  [d(X)]            
                                      
    [mark(c(X))] =  [2] X + [0]       
                 >= [1] X + [0]       
                 =  [a__c(X)]         
                                      
    [mark(h(X))] =  [2] X + [4]       
                 >  [2] X + [3]       
                 =  [a__h(mark(X))]   
                                      

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:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(X))))
  , a__c(X) -> d(X)
  , a__c(X) -> c(X)
  , mark(f(X)) -> a__f(mark(X)) }
Weak Trs:
  { a__h(X) -> a__c(d(X))
  , a__h(X) -> h(X)
  , mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(X)
  , mark(h(X)) -> a__h(mark(X)) }
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: { mark(f(X)) -> a__f(mark(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).
  
    [a__f](x1) = [1] x1 + [1]
                             
       [f](x1) = [1] x1 + [1]
                             
    [a__c](x1) = [1] x1 + [0]
                             
       [g](x1) = [1] x1 + [0]
                             
       [d](x1) = [1] x1 + [0]
                             
    [a__h](x1) = [1] x1 + [0]
                             
    [mark](x1) = [2] x1 + [1]
                             
       [c](x1) = [1] x1 + [0]
                             
       [h](x1) = [1] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X)] =  [1] X + [1]       
                 >= [1] X + [1]       
                 =  [f(X)]            
                                      
    [a__f(f(X))] =  [1] X + [2]       
                 >= [1] X + [2]       
                 =  [a__c(f(g(f(X))))]
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [d(X)]            
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [c(X)]            
                                      
       [a__h(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [a__c(d(X))]      
                                      
       [a__h(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [h(X)]            
                                      
    [mark(f(X))] =  [2] X + [3]       
                 >  [2] X + [2]       
                 =  [a__f(mark(X))]   
                                      
    [mark(g(X))] =  [2] X + [1]       
                 >  [1] X + [0]       
                 =  [g(X)]            
                                      
    [mark(d(X))] =  [2] X + [1]       
                 >  [1] X + [0]       
                 =  [d(X)]            
                                      
    [mark(c(X))] =  [2] X + [1]       
                 >  [1] X + [0]       
                 =  [a__c(X)]         
                                      
    [mark(h(X))] =  [2] X + [1]       
                 >= [2] X + [1]       
                 =  [a__h(mark(X))]   
                                      

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:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(X))))
  , a__c(X) -> d(X)
  , a__c(X) -> c(X) }
Weak Trs:
  { a__h(X) -> a__c(d(X))
  , a__h(X) -> h(X)
  , mark(f(X)) -> a__f(mark(X))
  , mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(X)
  , mark(h(X)) -> a__h(mark(X)) }
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:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(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).
  
    [a__f](x1) = [1] x1 + [2]
                             
       [f](x1) = [1] x1 + [1]
                             
    [a__c](x1) = [1] x1 + [0]
                             
       [g](x1) = [1] x1 + [0]
                             
       [d](x1) = [1] x1 + [0]
                             
    [a__h](x1) = [1] x1 + [0]
                             
    [mark](x1) = [3] x1 + [1]
                             
       [c](x1) = [1] x1 + [0]
                             
       [h](x1) = [1] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X)] =  [1] X + [2]       
                 >  [1] X + [1]       
                 =  [f(X)]            
                                      
    [a__f(f(X))] =  [1] X + [3]       
                 >  [1] X + [2]       
                 =  [a__c(f(g(f(X))))]
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [d(X)]            
                                      
       [a__c(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [c(X)]            
                                      
       [a__h(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [a__c(d(X))]      
                                      
       [a__h(X)] =  [1] X + [0]       
                 >= [1] X + [0]       
                 =  [h(X)]            
                                      
    [mark(f(X))] =  [3] X + [4]       
                 >  [3] X + [3]       
                 =  [a__f(mark(X))]   
                                      
    [mark(g(X))] =  [3] X + [1]       
                 >  [1] X + [0]       
                 =  [g(X)]            
                                      
    [mark(d(X))] =  [3] X + [1]       
                 >  [1] X + [0]       
                 =  [d(X)]            
                                      
    [mark(c(X))] =  [3] X + [1]       
                 >  [1] X + [0]       
                 =  [a__c(X)]         
                                      
    [mark(h(X))] =  [3] X + [1]       
                 >= [3] X + [1]       
                 =  [a__h(mark(X))]   
                                      

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:
  { a__c(X) -> d(X)
  , a__c(X) -> c(X) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(X))))
  , a__h(X) -> a__c(d(X))
  , a__h(X) -> h(X)
  , mark(f(X)) -> a__f(mark(X))
  , mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(X)
  , mark(h(X)) -> a__h(mark(X)) }
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: { a__c(X) -> d(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).
  
    [a__f](x1) = [1] x1 + [3]
                             
       [f](x1) = [1] x1 + [1]
                             
    [a__c](x1) = [1] x1 + [2]
                             
       [g](x1) = [1] x1 + [0]
                             
       [d](x1) = [1] x1 + [0]
                             
    [a__h](x1) = [1] x1 + [2]
                             
    [mark](x1) = [3] x1 + [0]
                             
       [c](x1) = [1] x1 + [2]
                             
       [h](x1) = [1] x1 + [2]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X)] =  [1] X + [3]       
                 >  [1] X + [1]       
                 =  [f(X)]            
                                      
    [a__f(f(X))] =  [1] X + [4]       
                 >= [1] X + [4]       
                 =  [a__c(f(g(f(X))))]
                                      
       [a__c(X)] =  [1] X + [2]       
                 >  [1] X + [0]       
                 =  [d(X)]            
                                      
       [a__c(X)] =  [1] X + [2]       
                 >= [1] X + [2]       
                 =  [c(X)]            
                                      
       [a__h(X)] =  [1] X + [2]       
                 >= [1] X + [2]       
                 =  [a__c(d(X))]      
                                      
       [a__h(X)] =  [1] X + [2]       
                 >= [1] X + [2]       
                 =  [h(X)]            
                                      
    [mark(f(X))] =  [3] X + [3]       
                 >= [3] X + [3]       
                 =  [a__f(mark(X))]   
                                      
    [mark(g(X))] =  [3] X + [0]       
                 >= [1] X + [0]       
                 =  [g(X)]            
                                      
    [mark(d(X))] =  [3] X + [0]       
                 >= [1] X + [0]       
                 =  [d(X)]            
                                      
    [mark(c(X))] =  [3] X + [6]       
                 >  [1] X + [2]       
                 =  [a__c(X)]         
                                      
    [mark(h(X))] =  [3] X + [6]       
                 >  [3] X + [2]       
                 =  [a__h(mark(X))]   
                                      

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: { a__c(X) -> c(X) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(X))))
  , a__c(X) -> d(X)
  , a__h(X) -> a__c(d(X))
  , a__h(X) -> h(X)
  , mark(f(X)) -> a__f(mark(X))
  , mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(X)
  , mark(h(X)) -> a__h(mark(X)) }
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: { a__c(X) -> c(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).
  
    [a__f](x1) = [1] x1 + [3]
                             
       [f](x1) = [1] x1 + [2]
                             
    [a__c](x1) = [1] x1 + [1]
                             
       [g](x1) = [1] x1 + [0]
                             
       [d](x1) = [1] x1 + [0]
                             
    [a__h](x1) = [1] x1 + [1]
                             
    [mark](x1) = [2] x1 + [2]
                             
       [c](x1) = [1] x1 + [0]
                             
       [h](x1) = [1] x1 + [1]
  
  This order satisfies the following ordering constraints:
  
       [a__f(X)] =  [1] X + [3]       
                 >  [1] X + [2]       
                 =  [f(X)]            
                                      
    [a__f(f(X))] =  [1] X + [5]       
                 >= [1] X + [5]       
                 =  [a__c(f(g(f(X))))]
                                      
       [a__c(X)] =  [1] X + [1]       
                 >  [1] X + [0]       
                 =  [d(X)]            
                                      
       [a__c(X)] =  [1] X + [1]       
                 >  [1] X + [0]       
                 =  [c(X)]            
                                      
       [a__h(X)] =  [1] X + [1]       
                 >= [1] X + [1]       
                 =  [a__c(d(X))]      
                                      
       [a__h(X)] =  [1] X + [1]       
                 >= [1] X + [1]       
                 =  [h(X)]            
                                      
    [mark(f(X))] =  [2] X + [6]       
                 >  [2] X + [5]       
                 =  [a__f(mark(X))]   
                                      
    [mark(g(X))] =  [2] X + [2]       
                 >  [1] X + [0]       
                 =  [g(X)]            
                                      
    [mark(d(X))] =  [2] X + [2]       
                 >  [1] X + [0]       
                 =  [d(X)]            
                                      
    [mark(c(X))] =  [2] X + [2]       
                 >  [1] X + [1]       
                 =  [a__c(X)]         
                                      
    [mark(h(X))] =  [2] X + [4]       
                 >  [2] X + [3]       
                 =  [a__h(mark(X))]   
                                      

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:
  { a__f(X) -> f(X)
  , a__f(f(X)) -> a__c(f(g(f(X))))
  , a__c(X) -> d(X)
  , a__c(X) -> c(X)
  , a__h(X) -> a__c(d(X))
  , a__h(X) -> h(X)
  , mark(f(X)) -> a__f(mark(X))
  , mark(g(X)) -> g(X)
  , mark(d(X)) -> d(X)
  , mark(c(X)) -> a__c(X)
  , mark(h(X)) -> a__h(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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