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(a())) -> a__f(g(f(a())))
  , mark(f(X)) -> a__f(X)
  , mark(a()) -> a()
  , mark(g(X)) -> g(mark(X)) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

The input is overlay and right-linear. Switching to innermost
rewriting.

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(a())) -> a__f(g(f(a())))
  , mark(f(X)) -> a__f(X)
  , mark(a()) -> a()
  , mark(g(X)) -> g(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add the following weak dependency pairs:

Strict DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(f(a())) -> c_2(a__f^#(g(f(a()))))
  , mark^#(f(X)) -> c_3(a__f^#(X))
  , mark^#(a()) -> c_4()
  , mark^#(g(X)) -> c_5(mark^#(X)) }

and mark the set of starting terms.

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

Strict DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(f(a())) -> c_2(a__f^#(g(f(a()))))
  , mark^#(f(X)) -> c_3(a__f^#(X))
  , mark^#(a()) -> c_4()
  , mark^#(g(X)) -> c_5(mark^#(X)) }
Strict Trs:
  { a__f(X) -> f(X)
  , a__f(f(a())) -> a__f(g(f(a())))
  , mark(f(X)) -> a__f(X)
  , mark(a()) -> a()
  , mark(g(X)) -> g(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

No rule is usable, rules are removed from the input problem.

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

Strict DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(f(a())) -> c_2(a__f^#(g(f(a()))))
  , mark^#(f(X)) -> c_3(a__f^#(X))
  , mark^#(a()) -> c_4()
  , mark^#(g(X)) -> c_5(mark^#(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following constant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(c_2) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1}

TcT has computed the following constructor-restricted matrix
interpretation.

       [f](x1) = [0]           
                 [0]           
                               
           [a] = [0]           
                 [0]           
                               
       [g](x1) = [0]           
                 [0]           
                               
  [a__f^#](x1) = [0]           
                 [0]           
                               
         [c_1] = [0]           
                 [0]           
                               
     [c_2](x1) = [1 0] x1 + [0]
                 [0 1]      [0]
                               
  [mark^#](x1) = [1]           
                 [0]           
                               
     [c_3](x1) = [1 0] x1 + [0]
                 [0 1]      [0]
                               
         [c_4] = [0]           
                 [0]           
                               
     [c_5](x1) = [1 0] x1 + [0]
                 [0 1]      [0]

The order satisfies the following ordering constraints:

       [a__f^#(X)] =  [0]                     
                      [0]                     
                   >= [0]                     
                      [0]                     
                   =  [c_1()]                 
                                              
  [a__f^#(f(a()))] =  [0]                     
                      [0]                     
                   >= [0]                     
                      [0]                     
                   =  [c_2(a__f^#(g(f(a()))))]
                                              
    [mark^#(f(X))] =  [1]                     
                      [0]                     
                   >  [0]                     
                      [0]                     
                   =  [c_3(a__f^#(X))]        
                                              
     [mark^#(a())] =  [1]                     
                      [0]                     
                   >  [0]                     
                      [0]                     
                   =  [c_4()]                 
                                              
    [mark^#(g(X))] =  [1]                     
                      [0]                     
                   >= [1]                     
                      [0]                     
                   =  [c_5(mark^#(X))]        
                                              

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

Strict DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(f(a())) -> c_2(a__f^#(g(f(a()))))
  , mark^#(g(X)) -> c_5(mark^#(X)) }
Weak DPs:
  { mark^#(f(X)) -> c_3(a__f^#(X))
  , mark^#(a()) -> c_4() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ mark^#(a()) -> c_4() }

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

Strict DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(f(a())) -> c_2(a__f^#(g(f(a()))))
  , mark^#(g(X)) -> c_5(mark^#(X)) }
Weak DPs: { mark^#(f(X)) -> c_3(a__f^#(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.

DPs:
  { 1: a__f^#(X) -> c_1()
  , 4: mark^#(f(X)) -> c_3(a__f^#(X)) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_2) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
         [f](x1) = [0]         
                               
             [a] = [0]         
                               
         [g](x1) = [0]         
                               
    [a__f^#](x1) = [1]         
                               
           [c_1] = [0]         
                               
       [c_2](x1) = [1] x1 + [0]
                               
    [mark^#](x1) = [4]         
                               
       [c_3](x1) = [3] x1 + [0]
                               
       [c_5](x1) = [1] x1 + [0]
  
  The order satisfies the following ordering constraints:
  
         [a__f^#(X)] =  [1]                     
                     >  [0]                     
                     =  [c_1()]                 
                                                
    [a__f^#(f(a()))] =  [1]                     
                     >= [1]                     
                     =  [c_2(a__f^#(g(f(a()))))]
                                                
      [mark^#(f(X))] =  [4]                     
                     >  [3]                     
                     =  [c_3(a__f^#(X))]        
                                                
      [mark^#(g(X))] =  [4]                     
                     >= [4]                     
                     =  [c_5(mark^#(X))]        
                                                

We return to the main proof. Consider the set of all dependency
pairs

:
  { 1: a__f^#(X) -> c_1()
  , 2: a__f^#(f(a())) -> c_2(a__f^#(g(f(a()))))
  , 3: mark^#(g(X)) -> c_5(mark^#(X))
  , 4: mark^#(f(X)) -> c_3(a__f^#(X)) }

Processor 'matrix interpretation of dimension 1' induces the
complexity certificate YES(?,O(n^1)) on application of dependency
pairs {1,4}. These cover all (indirect) predecessors of dependency
pairs {1,2,4}, their number of application is equally bounded. The
dependency pairs are shifted into the weak component.

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

Strict DPs: { mark^#(g(X)) -> c_5(mark^#(X)) }
Weak DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(f(a())) -> c_2(a__f^#(g(f(a()))))
  , mark^#(f(X)) -> c_3(a__f^#(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ a__f^#(X) -> c_1()
, a__f^#(f(a())) -> c_2(a__f^#(g(f(a()))))
, mark^#(f(X)) -> c_3(a__f^#(X)) }

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

Strict DPs: { mark^#(g(X)) -> c_5(mark^#(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

DPs:
  { 1: mark^#(g(X)) -> c_5(mark^#(X)) }

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(f) = {1}, safe(a) = {}, safe(g) = {1}, safe(a__f^#) = {},
   safe(c_1) = {}, safe(c_2) = {}, safe(mark^#) = {}, safe(c_3) = {},
   safe(c_5) = {}
  
  and precedence
  
   empty .
  
  Following symbols are considered recursive:
  
   {mark^#}
  
  The recursion depth is 1.
  
  Further, following argument filtering is employed:
  
   pi(f) = [], pi(a) = [], pi(g) = [1], pi(a__f^#) = [], pi(c_1) = [],
   pi(c_2) = [], pi(mark^#) = [1], pi(c_3) = [], pi(c_5) = [1]
  
  Usable defined function symbols are a subset of:
  
   {a__f^#, mark^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
    pi(mark^#(g(X))) = mark^#(g(; X);)   
                     > c_5(mark^#(X;);)  
                     = pi(c_5(mark^#(X)))
                                         

The strictly oriented rules are moved into the weak component.

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

Weak DPs: { mark^#(g(X)) -> c_5(mark^#(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ mark^#(g(X)) -> c_5(mark^#(X)) }

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

Rules: Empty
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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