YES(O(1),O(n^3))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict Trs:
  { f(s(x1)) -> s(s(f(p(s(x1)))))
  , f(0(x1)) -> 0(x1)
  , p(s(x1)) -> x1 }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We add following weak dependency pairs:

Strict DPs:
  { f^#(s(x1)) -> c_1(f^#(p(s(x1))))
  , f^#(0(x1)) -> c_2()
  , p^#(s(x1)) -> 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^3)).

Strict DPs:
  { f^#(s(x1)) -> c_1(f^#(p(s(x1))))
  , f^#(0(x1)) -> c_2()
  , p^#(s(x1)) -> c_3() }
Strict Trs:
  { f(s(x1)) -> s(s(f(p(s(x1)))))
  , f(0(x1)) -> 0(x1)
  , p(s(x1)) -> x1 }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We replace rewrite rules by usable rules:

  Strict Usable Rules: { p(s(x1)) -> x1 }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { f^#(s(x1)) -> c_1(f^#(p(s(x1))))
  , f^#(0(x1)) -> c_2()
  , p^#(s(x1)) -> c_3() }
Strict Trs: { p(s(x1)) -> x1 }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

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

The following argument positions are usable:
  Uargs(f^#) = {1}, Uargs(c_1) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

    [s](x1) = [1] x1 + [2]
                          
    [p](x1) = [1] x1 + [0]
                          
    [0](x1) = [1]         
                          
  [f^#](x1) = [2] x1 + [2]
                          
  [c_1](x1) = [1] x1 + [1]
                          
      [c_2] = [1]         
                          
  [p^#](x1) = [1] x1 + [1]
                          
      [c_3] = [2]         

This order satisfies following ordering constraints:

  [p(s(x1))] = [1] x1 + [2]
             > [1] x1 + [0]
             = [x1]        
                           

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(n^3)).

Strict DPs: { f^#(s(x1)) -> c_1(f^#(p(s(x1)))) }
Weak DPs:
  { f^#(0(x1)) -> c_2()
  , p^#(s(x1)) -> c_3() }
Weak Trs: { p(s(x1)) -> x1 }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

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

{ f^#(0(x1)) -> c_2()
, p^#(s(x1)) -> c_3() }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict DPs: { f^#(s(x1)) -> c_1(f^#(p(s(x1)))) }
Weak Trs: { p(s(x1)) -> x1 }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

The following argument positions are usable:
  Uargs(c_1) = {1}

TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).

              [1 0 1]      [0]
    [s](x1) = [0 1 0] x1 + [1]
              [0 1 0]      [0]
                              
              [1 0 0]      [0]
    [p](x1) = [0 0 1] x1 + [0]
              [1 0 0]      [0]
                              
              [0 1 0]      [0]
  [f^#](x1) = [0 0 0] x1 + [0]
              [0 0 0]      [0]
                              
              [1 0 0]      [0]
  [c_1](x1) = [0 0 0] x1 + [0]
              [0 0 0]      [0]

This order satisfies following ordering constraints:

    [p(s(x1))] =  [1 0 1]      [0]    
                  [0 1 0] x1 + [0]    
                  [1 0 1]      [0]    
               >= [1 0 0]      [0]    
                  [0 1 0] x1 + [0]    
                  [0 0 1]      [0]    
               =  [x1]                
                                      
  [f^#(s(x1))] =  [0 1 0]      [1]    
                  [0 0 0] x1 + [0]    
                  [0 0 0]      [0]    
               >  [0 1 0]      [0]    
                  [0 0 0] x1 + [0]    
                  [0 0 0]      [0]    
               =  [c_1(f^#(p(s(x1))))]
                                      

Hurray, we answered YES(O(1),O(n^3))