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

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

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

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__g](x1) = [2] x1 + [1]
                             
    [a__h](x1) = [2] x1 + [0]
                             
        [a__c] = [2]         
                             
           [d] = [2]         
                             
           [c] = [1]         
                             
    [mark](x1) = [2] x1 + [0]
                             
       [g](x1) = [1] x1 + [1]
                             
       [h](x1) = [1] x1 + [0]
  
  This order satisfies the following ordering constraints:
  
       [a__g(X)] =  [2] X + [1]
                 >  [2] X + [0]
                 =  [a__h(X)]  
                               
       [a__g(X)] =  [2] X + [1]
                 >= [1] X + [1]
                 =  [g(X)]     
                               
       [a__h(X)] =  [2] X + [0]
                 >= [1] X + [0]
                 =  [h(X)]     
                               
     [a__h(d())] =  [4]        
                 >  [3]        
                 =  [a__g(c())]
                               
        [a__c()] =  [2]        
                 >= [2]        
                 =  [d()]      
                               
        [a__c()] =  [2]        
                 >  [1]        
                 =  [c()]      
                               
     [mark(d())] =  [4]        
                 >  [2]        
                 =  [d()]      
                               
     [mark(c())] =  [2]        
                 >= [2]        
                 =  [a__c()]   
                               
    [mark(g(X))] =  [2] X + [2]
                 >  [2] X + [1]
                 =  [a__g(X)]  
                               
    [mark(h(X))] =  [2] X + [0]
                 >= [2] X + [0]
                 =  [a__h(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__g(X) -> a__h(X)
  , a__g(X) -> g(X)
  , a__h(X) -> h(X)
  , a__h(d()) -> a__g(c())
  , a__c() -> d()
  , a__c() -> c()
  , mark(d()) -> d()
  , mark(c()) -> a__c()
  , mark(g(X)) -> a__g(X)
  , mark(h(X)) -> a__h(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))