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:
  { fac(s(x)) -> *(fac(p(s(x))), s(x))
  , p(s(s(x))) -> s(p(s(x)))
  , p(s(0())) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We add the following dependency tuples:

Strict DPs:
  { fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x)))
  , p^#(s(s(x))) -> c_2(p^#(s(x)))
  , p^#(s(0())) -> c_3() }

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:
  { fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x)))
  , p^#(s(s(x))) -> c_2(p^#(s(x)))
  , p^#(s(0())) -> c_3() }
Weak Trs:
  { fac(s(x)) -> *(fac(p(s(x))), s(x))
  , p(s(s(x))) -> s(p(s(x)))
  , p(s(0())) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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

  DPs:
    { 1: fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x)))
    , 2: p^#(s(s(x))) -> c_2(p^#(s(x)))
    , 3: p^#(s(0())) -> c_3() }

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

Strict DPs:
  { fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x)))
  , p^#(s(s(x))) -> c_2(p^#(s(x))) }
Weak DPs: { p^#(s(0())) -> c_3() }
Weak Trs:
  { fac(s(x)) -> *(fac(p(s(x))), s(x))
  , p(s(s(x))) -> s(p(s(x)))
  , p(s(0())) -> 0() }
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.

{ p^#(s(0())) -> c_3() }

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

Strict DPs:
  { fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x)))
  , p^#(s(s(x))) -> c_2(p^#(s(x))) }
Weak Trs:
  { fac(s(x)) -> *(fac(p(s(x))), s(x))
  , p(s(s(x))) -> s(p(s(x)))
  , p(s(0())) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { p(s(s(x))) -> s(p(s(x)))
    , p(s(0())) -> 0() }

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

Strict DPs:
  { fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x)))
  , p^#(s(s(x))) -> c_2(p^#(s(x))) }
Weak Trs:
  { p(s(s(x))) -> s(p(s(x)))
  , p(s(0())) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We decompose the input problem according to the dependency graph
into the upper component

  { fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x))) }

and lower component

  { p^#(s(s(x))) -> c_2(p^#(s(x))) }

Further, following extension rules are added to the lower
component.

{ fac^#(s(x)) -> fac^#(p(s(x)))
, fac^#(s(x)) -> p^#(s(x)) }

TcT solves the upper component with certificate YES(O(1),O(n^1)).

Sub-proof:
----------
  We are left with following problem, upon which TcT provides the
  certificate YES(O(1),O(n^1)).
  
  Strict DPs: { fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x))) }
  Weak Trs:
    { p(s(s(x))) -> s(p(s(x)))
    , p(s(0())) -> 0() }
  Obligation:
    innermost runtime complexity
  Answer:
    YES(O(1),O(n^1))
  
  We use the processor 'matrix interpretation of dimension 2' to
  orient following rules strictly.
  
  DPs:
    { 1: fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x))) }
  
  Sub-proof:
  ----------
    The following argument positions are usable:
      Uargs(c_1) = {1}
    
    TcT has computed the following constructor-based matrix
    interpretation satisfying not(EDA) and not(IDA(1)).
    
          [fac](x1) = [7 7] x1 + [0]
                      [0 0]      [0]
                                    
            [s](x1) = [1 0] x1 + [1]
                      [1 0]      [0]
                                    
        [*](x1, x2) = [0]           
                      [0]           
                                    
            [p](x1) = [0 1] x1 + [0]
                      [2 0]      [0]
                                    
                [0] = [0]           
                      [0]           
                                    
        [fac^#](x1) = [1 0] x1 + [0]
                      [4 0]      [0]
                                    
      [c_1](x1, x2) = [1 0] x1 + [0]
                      [0 0]      [3]
                                    
          [p^#](x1) = [0]           
                      [0]           
                                    
          [c_2](x1) = [7 7] x1 + [0]
                      [0 0]      [0]
                                    
              [c_3] = [0]           
                      [0]           
    
    The following symbols are considered usable
    
      {p, fac^#}
    
    The order satisfies the following ordering constraints:
    
       [p(s(s(x)))] =  [1 0] x + [1]                   
                       [2 0]     [4]                   
                    >= [1 0] x + [1]                   
                       [1 0]     [0]                   
                    =  [s(p(s(x)))]                    
                                                       
        [p(s(0()))] =  [0]                             
                       [2]                             
                    >= [0]                             
                       [0]                             
                    =  [0()]                           
                                                       
      [fac^#(s(x))] =  [1 0] x + [1]                   
                       [4 0]     [4]                   
                    >  [1 0] x + [0]                   
                       [0 0]     [3]                   
                    =  [c_1(fac^#(p(s(x))), p^#(s(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: { fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x))) }
  Weak Trs:
    { p(s(s(x))) -> s(p(s(x)))
    , p(s(0())) -> 0() }
  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.
  
  { fac^#(s(x)) -> c_1(fac^#(p(s(x))), p^#(s(x))) }
  
  We are left with following problem, upon which TcT provides the
  certificate YES(O(1),O(1)).
  
  Weak Trs:
    { p(s(s(x))) -> s(p(s(x)))
    , p(s(0())) -> 0() }
  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

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 DPs: { p^#(s(s(x))) -> c_2(p^#(s(x))) }
Weak DPs:
  { fac^#(s(x)) -> fac^#(p(s(x)))
  , fac^#(s(x)) -> p^#(s(x)) }
Weak Trs:
  { p(s(s(x))) -> s(p(s(x)))
  , p(s(0())) -> 0() }
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: p^#(s(s(x))) -> c_2(p^#(s(x)))
  , 3: fac^#(s(x)) -> p^#(s(x)) }
Trs: { p(s(0())) -> 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(fac) = {}, safe(s) = {1}, safe(*) = {1, 2}, safe(p) = {1},
   safe(0) = {}, safe(fac^#) = {}, safe(c_1) = {}, safe(p^#) = {},
   safe(c_2) = {}, safe(c_3) = {}
  
  and precedence
  
   fac^# > p, fac^# > p^#, p ~ p^# .
  
  Following symbols are considered recursive:
  
   {p, p^#}
  
  The recursion depth is 1.
  
  Further, following argument filtering is employed:
  
   pi(fac) = [], pi(s) = [1], pi(*) = [], pi(p) = 1, pi(0) = [],
   pi(fac^#) = [1], pi(c_1) = [], pi(p^#) = [1], pi(c_2) = [1],
   pi(c_3) = []
  
  Usable defined function symbols are a subset of:
  
   {p, fac^#, p^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
     pi(fac^#(s(x))) =  fac^#(s(; x);)    
                     >= fac^#(s(; x);)    
                     =  pi(fac^#(p(s(x))))
                                          
     pi(fac^#(s(x))) =  fac^#(s(; x);)    
                     >  p^#(s(; x);)      
                     =  pi(p^#(s(x)))     
                                          
    pi(p^#(s(s(x)))) =  p^#(s(; s(; x));) 
                     >  c_2(p^#(s(; x););)
                     =  pi(c_2(p^#(s(x))))
                                          
      pi(p(s(s(x)))) =  s(; s(; x))       
                     >= s(; s(; x))       
                     =  pi(s(p(s(x))))    
                                          
       pi(p(s(0()))) =  s(; 0())          
                     >  0()               
                     =  pi(0())           
                                          

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:
  { fac^#(s(x)) -> fac^#(p(s(x)))
  , fac^#(s(x)) -> p^#(s(x))
  , p^#(s(s(x))) -> c_2(p^#(s(x))) }
Weak Trs:
  { p(s(s(x))) -> s(p(s(x)))
  , p(s(0())) -> 0() }
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.

{ fac^#(s(x)) -> fac^#(p(s(x)))
, fac^#(s(x)) -> p^#(s(x))
, p^#(s(s(x))) -> c_2(p^#(s(x))) }

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

Weak Trs:
  { p(s(s(x))) -> s(p(s(x)))
  , p(s(0())) -> 0() }
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

Hurray, we answered YES(O(1),O(n^2))