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:
  { f(0()) -> true()
  , f(1()) -> false()
  , f(s(x)) -> f(x)
  , if(true(), s(x), s(y)) -> s(x)
  , if(false(), s(x), s(y)) -> s(y)
  , g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
  , g(x, c(y)) -> c(g(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We add the following dependency tuples:

Strict DPs:
  { f^#(0()) -> c_1()
  , f^#(1()) -> c_2()
  , f^#(s(x)) -> c_3(f^#(x))
  , if^#(true(), s(x), s(y)) -> c_4()
  , if^#(false(), s(x), s(y)) -> c_5()
  , g^#(x, c(y)) ->
    c_6(g^#(x, if(f(x), c(g(s(x), y)), c(y))),
        if^#(f(x), c(g(s(x), y)), c(y)),
        f^#(x),
        g^#(s(x), y))
  , g^#(x, c(y)) -> c_7(g^#(x, y)) }

and mark the set of starting terms.

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

Strict DPs:
  { f^#(0()) -> c_1()
  , f^#(1()) -> c_2()
  , f^#(s(x)) -> c_3(f^#(x))
  , if^#(true(), s(x), s(y)) -> c_4()
  , if^#(false(), s(x), s(y)) -> c_5()
  , g^#(x, c(y)) ->
    c_6(g^#(x, if(f(x), c(g(s(x), y)), c(y))),
        if^#(f(x), c(g(s(x), y)), c(y)),
        f^#(x),
        g^#(s(x), y))
  , g^#(x, c(y)) -> c_7(g^#(x, y)) }
Weak Trs:
  { f(0()) -> true()
  , f(1()) -> false()
  , f(s(x)) -> f(x)
  , if(true(), s(x), s(y)) -> s(x)
  , if(false(), s(x), s(y)) -> s(y)
  , g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
  , g(x, c(y)) -> c(g(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We estimate the number of application of {1,2,4,5} by applications
of Pre({1,2,4,5}) = {3,6}. Here rules are labeled as follows:

  DPs:
    { 1: f^#(0()) -> c_1()
    , 2: f^#(1()) -> c_2()
    , 3: f^#(s(x)) -> c_3(f^#(x))
    , 4: if^#(true(), s(x), s(y)) -> c_4()
    , 5: if^#(false(), s(x), s(y)) -> c_5()
    , 6: g^#(x, c(y)) ->
         c_6(g^#(x, if(f(x), c(g(s(x), y)), c(y))),
             if^#(f(x), c(g(s(x), y)), c(y)),
             f^#(x),
             g^#(s(x), y))
    , 7: g^#(x, c(y)) -> c_7(g^#(x, y)) }

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

Strict DPs:
  { f^#(s(x)) -> c_3(f^#(x))
  , g^#(x, c(y)) ->
    c_6(g^#(x, if(f(x), c(g(s(x), y)), c(y))),
        if^#(f(x), c(g(s(x), y)), c(y)),
        f^#(x),
        g^#(s(x), y))
  , g^#(x, c(y)) -> c_7(g^#(x, y)) }
Weak DPs:
  { f^#(0()) -> c_1()
  , f^#(1()) -> c_2()
  , if^#(true(), s(x), s(y)) -> c_4()
  , if^#(false(), s(x), s(y)) -> c_5() }
Weak Trs:
  { f(0()) -> true()
  , f(1()) -> false()
  , f(s(x)) -> f(x)
  , if(true(), s(x), s(y)) -> s(x)
  , if(false(), s(x), s(y)) -> s(y)
  , g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
  , g(x, c(y)) -> c(g(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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

{ f^#(0()) -> c_1()
, f^#(1()) -> c_2()
, if^#(true(), s(x), s(y)) -> c_4()
, if^#(false(), s(x), s(y)) -> c_5() }

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

Strict DPs:
  { f^#(s(x)) -> c_3(f^#(x))
  , g^#(x, c(y)) ->
    c_6(g^#(x, if(f(x), c(g(s(x), y)), c(y))),
        if^#(f(x), c(g(s(x), y)), c(y)),
        f^#(x),
        g^#(s(x), y))
  , g^#(x, c(y)) -> c_7(g^#(x, y)) }
Weak Trs:
  { f(0()) -> true()
  , f(1()) -> false()
  , f(s(x)) -> f(x)
  , if(true(), s(x), s(y)) -> s(x)
  , if(false(), s(x), s(y)) -> s(y)
  , g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
  , g(x, c(y)) -> c(g(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { g^#(x, c(y)) ->
    c_6(g^#(x, if(f(x), c(g(s(x), y)), c(y))),
        if^#(f(x), c(g(s(x), y)), c(y)),
        f^#(x),
        g^#(s(x), y)) }

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

Strict DPs:
  { f^#(s(x)) -> c_1(f^#(x))
  , g^#(x, c(y)) -> c_2(g^#(x, y))
  , g^#(x, c(y)) -> c_3(f^#(x), g^#(s(x), y)) }
Weak Trs:
  { f(0()) -> true()
  , f(1()) -> false()
  , f(s(x)) -> f(x)
  , if(true(), s(x), s(y)) -> s(x)
  , if(false(), s(x), s(y)) -> s(y)
  , g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
  , g(x, c(y)) -> c(g(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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^2)).

Strict DPs:
  { f^#(s(x)) -> c_1(f^#(x))
  , g^#(x, c(y)) -> c_2(g^#(x, y))
  , g^#(x, c(y)) -> c_3(f^#(x), g^#(s(x), y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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

DPs:
  { 1: f^#(s(x)) -> c_1(f^#(x)) }

Sub-proof:
----------
  The following argument positions are considered usable:
  
    Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1, 2}
  
  TcT has computed the following constructor-restricted polynomial
  interpretation.
  
                [f](x1) = 3*x1 + 3*x1^2                                                                                                            
                                                                                                                                                   
                  [0]() = 0                                                                                                                        
                                                                                                                                                   
               [true]() = 0                                                                                                                        
                                                                                                                                                   
                  [1]() = 0                                                                                                                        
                                                                                                                                                   
              [false]() = 0                                                                                                                        
                                                                                                                                                   
                [s](x1) = 1 + x1                                                                                                                   
                                                                                                                                                   
       [if](x1, x2, x3) = 3*x1 + 3*x1*x2 + 3*x1*x3 + 3*x1^2 + 3*x2 + 3*x2*x3 + 3*x2^2 + 3*x3 + 3*x3^2                                              
                                                                                                                                                   
            [g](x1, x2) = 3*x1 + 3*x1*x2 + 3*x1^2 + 3*x2 + 3*x2^2                                                                                  
                                                                                                                                                   
                [c](x1) = 1 + x1                                                                                                                   
                                                                                                                                                   
              [f^#](x1) = x1                                                                                                                       
                                                                                                                                                   
                [c_1]() = 0                                                                                                                        
                                                                                                                                                   
                [c_2]() = 0                                                                                                                        
                                                                                                                                                   
              [c_3](x1) = 3*x1 + 3*x1^2                                                                                                            
                                                                                                                                                   
     [if^#](x1, x2, x3) = 3*x1 + 3*x1*x2 + 3*x1*x3 + 3*x1^2 + 3*x2 + 3*x2*x3 + 3*x2^2 + 3*x3 + 3*x3^2                                              
                                                                                                                                                   
                [c_4]() = 0                                                                                                                        
                                                                                                                                                   
                [c_5]() = 0                                                                                                                        
                                                                                                                                                   
          [g^#](x1, x2) = x1*x2 + x2^2                                                                                                             
                                                                                                                                                   
  [c_6](x1, x2, x3, x4) = 3*x1 + 3*x1*x2 + 3*x1*x3 + 3*x1*x4 + 3*x1^2 + 3*x2 + 3*x2*x3 + 3*x2*x4 + 3*x2^2 + 3*x3 + 3*x3*x4 + 3*x3^2 + 3*x4 + 3*x4^2
                                                                                                                                                   
              [c_7](x1) = 3*x1 + 3*x1^2                                                                                                            
                                                                                                                                                   
                  [c]() = 0                                                                                                                        
                                                                                                                                                   
              [c_1](x1) = x1                                                                                                                       
                                                                                                                                                   
              [c_2](x1) = 1 + x1                                                                                                                   
                                                                                                                                                   
          [c_3](x1, x2) = 1 + x1 + x2                                                                                                              
                                                                                                                                                   
  
  The following symbols are considered usable
  
    {f^#, if^#, g^#}
  
  This order satisfies the following ordering constraints.
  
       [f^#(s(x))] =  1 + x                      
                   >  x                          
                   =  [c_1(f^#(x))]              
                                                 
    [g^#(x, c(y))] =  x + x*y + 1 + 2*y + y^2    
                   >= 1 + x*y + y^2              
                   =  [c_2(g^#(x, y))]           
                                                 
    [g^#(x, c(y))] =  x + x*y + 1 + 2*y + y^2    
                   >= 1 + x + y + x*y + y^2      
                   =  [c_3(f^#(x), g^#(s(x), y))]
                                                 

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(n^1)).

Strict DPs:
  { g^#(x, c(y)) -> c_2(g^#(x, y))
  , g^#(x, c(y)) -> c_3(f^#(x), g^#(s(x), y)) }
Weak DPs: { f^#(s(x)) -> c_1(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.

{ f^#(s(x)) -> c_1(f^#(x)) }

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

Strict DPs:
  { g^#(x, c(y)) -> c_2(g^#(x, y))
  , g^#(x, c(y)) -> c_3(f^#(x), g^#(s(x), y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { g^#(x, c(y)) -> c_3(f^#(x), g^#(s(x), y)) }

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

Strict DPs:
  { g^#(x, c(y)) -> c_1(g^#(x, y))
  , g^#(x, c(y)) -> c_2(g^#(s(x), y)) }
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: g^#(x, c(y)) -> c_1(g^#(x, y))
  , 2: g^#(x, c(y)) -> c_2(g^#(s(x), y)) }

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) = {}, safe(0) = {}, safe(true) = {}, safe(1) = {},
   safe(false) = {}, safe(s) = {1}, safe(if) = {}, safe(g) = {},
   safe(c) = {1}, safe(f^#) = {}, safe(c_1) = {}, safe(c_2) = {},
   safe(c_3) = {}, safe(if^#) = {}, safe(c_4) = {}, safe(c_5) = {},
   safe(g^#) = {1}, safe(c_6) = {}, safe(c_7) = {}, safe(c) = {},
   safe(c_1) = {}, safe(c_2) = {}, safe(c_3) = {}, safe(c) = {},
   safe(c_1) = {}, safe(c_2) = {}
  
  and precedence
  
   empty .
  
  Following symbols are considered recursive:
  
   {g^#}
  
  The recursion depth is 1.
  
  Further, following argument filtering is employed:
  
   pi(f) = [], pi(0) = [], pi(true) = [], pi(1) = [], pi(false) = [],
   pi(s) = 1, pi(if) = [], pi(g) = [], pi(c) = [1], pi(f^#) = [],
   pi(c_1) = [], pi(c_2) = [], pi(c_3) = [], pi(if^#) = [],
   pi(c_4) = [], pi(c_5) = [], pi(g^#) = [1, 2], pi(c_6) = [],
   pi(c_7) = [], pi(c) = [], pi(c_1) = [], pi(c_2) = [], pi(c_3) = [],
   pi(c) = [], pi(c_1) = [1], pi(c_2) = [1]
  
  Usable defined function symbols are a subset of:
  
   {f^#, if^#, g^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
    pi(g^#(x, c(y))) = g^#(c(; y); x)       
                     > c_1(g^#(y; x);)      
                     = pi(c_1(g^#(x, y)))   
                                            
    pi(g^#(x, c(y))) = g^#(c(; y); x)       
                     > c_2(g^#(y; x);)      
                     = pi(c_2(g^#(s(x), y)))
                                            

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:
  { g^#(x, c(y)) -> c_1(g^#(x, y))
  , g^#(x, c(y)) -> c_2(g^#(s(x), y)) }
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.

{ g^#(x, c(y)) -> c_1(g^#(x, y))
, g^#(x, c(y)) -> c_2(g^#(s(x), y)) }

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^2))