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

Strict Trs:
  { compS_f#1(compS_f(x2), x1) -> compS_f#1(x2, S(x1))
  , compS_f#1(id(), x3) -> S(x3)
  , iter#3(S(x6)) -> compS_f(iter#3(x6))
  , iter#3(0()) -> id()
  , main(S(x9)) -> compS_f#1(iter#3(x9), 0())
  , main(0()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add the following weak dependency pairs:

Strict DPs:
  { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
  , compS_f#1^#(id(), x3) -> c_2()
  , iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
  , iter#3^#(0()) -> c_4()
  , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0()))
  , main^#(0()) -> c_6() }

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:
  { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
  , compS_f#1^#(id(), x3) -> c_2()
  , iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
  , iter#3^#(0()) -> c_4()
  , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0()))
  , main^#(0()) -> c_6() }
Strict Trs:
  { compS_f#1(compS_f(x2), x1) -> compS_f#1(x2, S(x1))
  , compS_f#1(id(), x3) -> S(x3)
  , iter#3(S(x6)) -> compS_f(iter#3(x6))
  , iter#3(0()) -> id()
  , main(S(x9)) -> compS_f#1(iter#3(x9), 0())
  , main(0()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We replace rewrite rules by usable rules:

  Strict Usable Rules:
    { iter#3(S(x6)) -> compS_f(iter#3(x6))
    , iter#3(0()) -> id() }

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

Strict DPs:
  { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
  , compS_f#1^#(id(), x3) -> c_2()
  , iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
  , iter#3^#(0()) -> c_4()
  , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0()))
  , main^#(0()) -> c_6() }
Strict Trs:
  { iter#3(S(x6)) -> compS_f(iter#3(x6))
  , iter#3(0()) -> id() }
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(compS_f) = {1}, Uargs(compS_f#1^#) = {1}, Uargs(c_1) = {1},
  Uargs(c_3) = {1}, Uargs(c_5) = {1}

TcT has computed the following constructor-restricted matrix
interpretation.

          [compS_f](x1) = [1 0] x1 + [0]
                          [0 0]      [0]
                                        
                [S](x1) = [1 0] x1 + [2]
                          [0 1]      [1]
                                        
                   [id] = [0]           
                          [0]           
                                        
           [iter#3](x1) = [0 1] x1 + [0]
                          [2 0]      [0]
                                        
                    [0] = [0]           
                          [1]           
                                        
  [compS_f#1^#](x1, x2) = [1 0] x1 + [0]
                          [0 0]      [0]
                                        
              [c_1](x1) = [1 0] x1 + [0]
                          [0 1]      [0]
                                        
                  [c_2] = [0]           
                          [0]           
                                        
         [iter#3^#](x1) = [0]           
                          [0]           
                                        
              [c_3](x1) = [1 0] x1 + [0]
                          [0 1]      [0]
                                        
                  [c_4] = [0]           
                          [0]           
                                        
           [main^#](x1) = [0 1] x1 + [0]
                          [0 0]      [0]
                                        
              [c_5](x1) = [1 0] x1 + [0]
                          [0 1]      [0]
                                        
                  [c_6] = [0]           
                          [0]           

The order satisfies the following ordering constraints:

                 [iter#3(S(x6))] =  [0 1] x6 + [1]                     
                                    [2 0]      [4]                     
                                 >  [0 1] x6 + [0]                     
                                    [0 0]      [0]                     
                                 =  [compS_f(iter#3(x6))]              
                                                                       
                   [iter#3(0())] =  [1]                                
                                    [0]                                
                                 >  [0]                                
                                    [0]                                
                                 =  [id()]                             
                                                                       
  [compS_f#1^#(compS_f(x2), x1)] =  [1 0] x2 + [0]                     
                                    [0 0]      [0]                     
                                 >= [1 0] x2 + [0]                     
                                    [0 0]      [0]                     
                                 =  [c_1(compS_f#1^#(x2, S(x1)))]      
                                                                       
         [compS_f#1^#(id(), x3)] =  [0]                                
                                    [0]                                
                                 >= [0]                                
                                    [0]                                
                                 =  [c_2()]                            
                                                                       
               [iter#3^#(S(x6))] =  [0]                                
                                    [0]                                
                                 >= [0]                                
                                    [0]                                
                                 =  [c_3(iter#3^#(x6))]                
                                                                       
                 [iter#3^#(0())] =  [0]                                
                                    [0]                                
                                 >= [0]                                
                                    [0]                                
                                 =  [c_4()]                            
                                                                       
                 [main^#(S(x9))] =  [0 1] x9 + [1]                     
                                    [0 0]      [0]                     
                                 >  [0 1] x9 + [0]                     
                                    [0 0]      [0]                     
                                 =  [c_5(compS_f#1^#(iter#3(x9), 0()))]
                                                                       
                   [main^#(0())] =  [1]                                
                                    [0]                                
                                 >  [0]                                
                                    [0]                                
                                 =  [c_6()]                            
                                                                       

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:
  { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
  , compS_f#1^#(id(), x3) -> c_2()
  , iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
  , iter#3^#(0()) -> c_4() }
Weak DPs:
  { main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0()))
  , main^#(0()) -> c_6() }
Weak Trs:
  { iter#3(S(x6)) -> compS_f(iter#3(x6))
  , iter#3(0()) -> id() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We estimate the number of application of {4} by applications of
Pre({4}) = {3}. Here rules are labeled as follows:

  DPs:
    { 1: compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
    , 2: compS_f#1^#(id(), x3) -> c_2()
    , 3: iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
    , 4: iter#3^#(0()) -> c_4()
    , 5: main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0()))
    , 6: main^#(0()) -> c_6() }

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

Strict DPs:
  { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
  , compS_f#1^#(id(), x3) -> c_2()
  , iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
Weak DPs:
  { iter#3^#(0()) -> c_4()
  , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0()))
  , main^#(0()) -> c_6() }
Weak Trs:
  { iter#3(S(x6)) -> compS_f(iter#3(x6))
  , iter#3(0()) -> id() }
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.

{ iter#3^#(0()) -> c_4()
, main^#(0()) -> c_6() }

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

Strict DPs:
  { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
  , compS_f#1^#(id(), x3) -> c_2()
  , iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
Weak DPs: { main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
Weak Trs:
  { iter#3(S(x6)) -> compS_f(iter#3(x6))
  , iter#3(0()) -> id() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We analyse the complexity of following sub-problems (R) and (S).
Problem (S) is obtained from the input problem by shifting strict
rules from (R) into the weak component:

Problem (R):
------------
  Strict DPs: { compS_f#1^#(id(), x3) -> c_2() }
  Weak DPs:
    { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
    , iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
    , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
  Weak Trs:
    { iter#3(S(x6)) -> compS_f(iter#3(x6))
    , iter#3(0()) -> id() }
  StartTerms: basic terms
  Strategy: innermost

Problem (S):
------------
  Strict DPs:
    { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
    , iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
  Weak DPs:
    { compS_f#1^#(id(), x3) -> c_2()
    , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
  Weak Trs:
    { iter#3(S(x6)) -> compS_f(iter#3(x6))
    , iter#3(0()) -> id() }
  StartTerms: basic terms
  Strategy: innermost

Overall, the transformation results in the following sub-problem(s):

Generated new problems:
-----------------------
R) Strict DPs: { compS_f#1^#(id(), x3) -> c_2() }
   Weak DPs:
     { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
     , iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
     , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
   Weak Trs:
     { iter#3(S(x6)) -> compS_f(iter#3(x6))
     , iter#3(0()) -> id() }
   StartTerms: basic terms
   Strategy: innermost
   
   This problem was proven YES(O(1),O(n^1)).

S) Strict DPs:
     { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
     , iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
   Weak DPs:
     { compS_f#1^#(id(), x3) -> c_2()
     , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
   Weak Trs:
     { iter#3(S(x6)) -> compS_f(iter#3(x6))
     , iter#3(0()) -> id() }
   StartTerms: basic terms
   Strategy: innermost
   
   This problem was proven YES(O(1),O(n^1)).


Proofs for generated problems:
------------------------------
R) We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs: { compS_f#1^#(id(), x3) -> c_2() }
   Weak DPs:
     { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
     , iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
     , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
   Weak Trs:
     { iter#3(S(x6)) -> compS_f(iter#3(x6))
     , iter#3(0()) -> id() }
   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.
   
   { iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
   
   We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs: { compS_f#1^#(id(), x3) -> c_2() }
   Weak DPs:
     { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
     , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
   Weak Trs:
     { iter#3(S(x6)) -> compS_f(iter#3(x6))
     , iter#3(0()) -> id() }
   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: compS_f#1^#(id(), x3) -> c_2()
     , 2: compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
     , 3: main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
   
   Sub-proof:
   ----------
     The input was oriented with the instance of 'Small Polynomial Path
     Order (PS,1-bounded)' as induced by the safe mapping
     
      safe(compS_f) = {1}, safe(S) = {1}, safe(id) = {},
      safe(iter#3) = {}, safe(0) = {}, safe(compS_f#1^#) = {2},
      safe(c_1) = {}, safe(c_2) = {}, safe(main^#) = {}, safe(c_5) = {}
     
     and precedence
     
      compS_f#1^# ~ main^# .
     
     Following symbols are considered recursive:
     
      {compS_f#1^#, main^#}
     
     The recursion depth is 1.
     
     Further, following argument filtering is employed:
     
      pi(compS_f) = [1], pi(S) = [1], pi(id) = [], pi(iter#3) = 1,
      pi(0) = [], pi(compS_f#1^#) = [1, 2], pi(c_1) = [1], pi(c_2) = [],
      pi(main^#) = [1], pi(c_5) = [1]
     
     Usable defined function symbols are a subset of:
     
      {iter#3, compS_f#1^#, main^#}
     
     For your convenience, here are the satisfied ordering constraints:
     
       pi(compS_f#1^#(compS_f(x2), x1)) =  compS_f#1^#(compS_f(; x2); x1)       
                                        >  c_1(compS_f#1^#(x2; S(; x1));)       
                                        =  pi(c_1(compS_f#1^#(x2, S(x1))))      
                                                                                
              pi(compS_f#1^#(id(), x3)) =  compS_f#1^#(id(); x3)                
                                        >  c_2()                                
                                        =  pi(c_2())                            
                                                                                
                      pi(main^#(S(x9))) =  main^#(S(; x9);)                     
                                        >  c_5(compS_f#1^#(x9; 0());)           
                                        =  pi(c_5(compS_f#1^#(iter#3(x9), 0())))
                                                                                
                      pi(iter#3(S(x6))) =  S(; x6)                              
                                        >= compS_f(; x6)                        
                                        =  pi(compS_f(iter#3(x6)))              
                                                                                
                        pi(iter#3(0())) =  0()                                  
                                        >= id()                                 
                                        =  pi(id())                             
                                                                                
   
   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:
     { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
     , compS_f#1^#(id(), x3) -> c_2()
     , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
   Weak Trs:
     { iter#3(S(x6)) -> compS_f(iter#3(x6))
     , iter#3(0()) -> id() }
   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.
   
   { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
   , compS_f#1^#(id(), x3) -> c_2()
   , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
   
   We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(1)).
   
   Weak Trs:
     { iter#3(S(x6)) -> compS_f(iter#3(x6))
     , iter#3(0()) -> id() }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(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(1)).
   
   Rules: Empty
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(1))
   
   Empty rules are trivially bounded

S) We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs:
     { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
     , iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
   Weak DPs:
     { compS_f#1^#(id(), x3) -> c_2()
     , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
   Weak Trs:
     { iter#3(S(x6)) -> compS_f(iter#3(x6))
     , iter#3(0()) -> id() }
   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.
   
   { compS_f#1^#(id(), x3) -> c_2() }
   
   We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs:
     { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
     , iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
   Weak DPs: { main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
   Weak Trs:
     { iter#3(S(x6)) -> compS_f(iter#3(x6))
     , iter#3(0()) -> id() }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(n^1))
   
   We analyse the complexity of following sub-problems (R) and (S).
   Problem (S) is obtained from the input problem by shifting strict
   rules from (R) into the weak component:
   
   Problem (R):
   ------------
     Strict DPs:
       { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1))) }
     Weak DPs:
       { iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
       , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
     Weak Trs:
       { iter#3(S(x6)) -> compS_f(iter#3(x6))
       , iter#3(0()) -> id() }
     StartTerms: basic terms
     Strategy: innermost
   
   Problem (S):
   ------------
     Strict DPs: { iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
     Weak DPs:
       { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
       , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
     Weak Trs:
       { iter#3(S(x6)) -> compS_f(iter#3(x6))
       , iter#3(0()) -> id() }
     StartTerms: basic terms
     Strategy: innermost
   
   Overall, the transformation results in the following sub-problem(s):
   
   Generated new problems:
   -----------------------
   R) Strict DPs:
        { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1))) }
      Weak DPs:
        { iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
        , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
      Weak Trs:
        { iter#3(S(x6)) -> compS_f(iter#3(x6))
        , iter#3(0()) -> id() }
      StartTerms: basic terms
      Strategy: innermost
      
      This problem was proven YES(O(1),O(n^1)).
   
   S) Strict DPs: { iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
      Weak DPs:
        { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
        , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
      Weak Trs:
        { iter#3(S(x6)) -> compS_f(iter#3(x6))
        , iter#3(0()) -> id() }
      StartTerms: basic terms
      Strategy: innermost
      
      This problem was proven YES(O(1),O(n^1)).
   
   
   Proofs for generated problems:
   ------------------------------
   R) We are left with following problem, upon which TcT provides the
      certificate YES(O(1),O(n^1)).
      
      Strict DPs:
        { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1))) }
      Weak DPs:
        { iter#3^#(S(x6)) -> c_3(iter#3^#(x6))
        , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
      Weak Trs:
        { iter#3(S(x6)) -> compS_f(iter#3(x6))
        , iter#3(0()) -> id() }
      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.
      
      { iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
      
      We are left with following problem, upon which TcT provides the
      certificate YES(O(1),O(n^1)).
      
      Strict DPs:
        { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1))) }
      Weak DPs: { main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
      Weak Trs:
        { iter#3(S(x6)) -> compS_f(iter#3(x6))
        , iter#3(0()) -> id() }
      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: compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
        , 2: main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
      
      Sub-proof:
      ----------
        The input was oriented with the instance of 'Small Polynomial Path
        Order (PS,1-bounded)' as induced by the safe mapping
        
         safe(compS_f) = {1}, safe(S) = {1}, safe(id) = {},
         safe(iter#3) = {}, safe(0) = {}, safe(compS_f#1^#) = {2},
         safe(c_1) = {}, safe(main^#) = {}, safe(c_5) = {}
        
        and precedence
        
         compS_f#1^# ~ main^# .
        
        Following symbols are considered recursive:
        
         {compS_f#1^#, main^#}
        
        The recursion depth is 1.
        
        Further, following argument filtering is employed:
        
         pi(compS_f) = [1], pi(S) = [1], pi(id) = [], pi(iter#3) = 1,
         pi(0) = [], pi(compS_f#1^#) = [1], pi(c_1) = [1], pi(main^#) = [1],
         pi(c_5) = [1]
        
        Usable defined function symbols are a subset of:
        
         {iter#3, compS_f#1^#, main^#}
        
        For your convenience, here are the satisfied ordering constraints:
        
          pi(compS_f#1^#(compS_f(x2), x1)) =  compS_f#1^#(compS_f(; x2);)          
                                           >  c_1(compS_f#1^#(x2;);)               
                                           =  pi(c_1(compS_f#1^#(x2, S(x1))))      
                                                                                   
                         pi(main^#(S(x9))) =  main^#(S(; x9);)                     
                                           >  c_5(compS_f#1^#(x9;);)               
                                           =  pi(c_5(compS_f#1^#(iter#3(x9), 0())))
                                                                                   
                         pi(iter#3(S(x6))) =  S(; x6)                              
                                           >= compS_f(; x6)                        
                                           =  pi(compS_f(iter#3(x6)))              
                                                                                   
                           pi(iter#3(0())) =  0()                                  
                                           >= id()                                 
                                           =  pi(id())                             
                                                                                   
      
      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:
        { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
        , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
      Weak Trs:
        { iter#3(S(x6)) -> compS_f(iter#3(x6))
        , iter#3(0()) -> id() }
      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.
      
      { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
      , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
      
      We are left with following problem, upon which TcT provides the
      certificate YES(O(1),O(1)).
      
      Weak Trs:
        { iter#3(S(x6)) -> compS_f(iter#3(x6))
        , iter#3(0()) -> id() }
      Obligation:
        innermost runtime complexity
      Answer:
        YES(O(1),O(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(1)).
      
      Rules: Empty
      Obligation:
        innermost runtime complexity
      Answer:
        YES(O(1),O(1))
      
      Empty rules are trivially bounded
   
   S) We are left with following problem, upon which TcT provides the
      certificate YES(O(1),O(n^1)).
      
      Strict DPs: { iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
      Weak DPs:
        { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
        , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
      Weak Trs:
        { iter#3(S(x6)) -> compS_f(iter#3(x6))
        , iter#3(0()) -> id() }
      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.
      
      { compS_f#1^#(compS_f(x2), x1) -> c_1(compS_f#1^#(x2, S(x1)))
      , main^#(S(x9)) -> c_5(compS_f#1^#(iter#3(x9), 0())) }
      
      We are left with following problem, upon which TcT provides the
      certificate YES(O(1),O(n^1)).
      
      Strict DPs: { iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
      Weak Trs:
        { iter#3(S(x6)) -> compS_f(iter#3(x6))
        , iter#3(0()) -> id() }
      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: { iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
      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: iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
      
      Sub-proof:
      ----------
        The input was oriented with the instance of 'Small Polynomial Path
        Order (PS,1-bounded)' as induced by the safe mapping
        
         safe(S) = {1}, safe(iter#3^#) = {}, safe(c_3) = {}
        
        and precedence
        
         empty .
        
        Following symbols are considered recursive:
        
         {iter#3^#}
        
        The recursion depth is 1.
        
        Further, following argument filtering is employed:
        
         pi(S) = [1], pi(iter#3^#) = [1], pi(c_3) = [1]
        
        Usable defined function symbols are a subset of:
        
         {iter#3^#}
        
        For your convenience, here are the satisfied ordering constraints:
        
          pi(iter#3^#(S(x6))) = iter#3^#(S(; x6);)   
                              > c_3(iter#3^#(x6;);)  
                              = pi(c_3(iter#3^#(x6)))
                                                     
      
      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: { iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
      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.
      
      { iter#3^#(S(x6)) -> c_3(iter#3^#(x6)) }
      
      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))