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

We add following weak dependency pairs:

Strict DPs:
  { p^#(0()) -> c_1()
  , p^#(s(x)) -> c_2()
  , minus^#(x, 0()) -> c_3()
  , minus^#(x, s(y)) -> c_4(minus^#(p(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:
  { p^#(0()) -> c_1()
  , p^#(s(x)) -> c_2()
  , minus^#(x, 0()) -> c_3()
  , minus^#(x, s(y)) -> c_4(minus^#(p(x), y)) }
Strict Trs:
  { p(0()) -> 0()
  , p(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> minus(p(x), y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We replace rewrite rules by usable rules:

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

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

Strict DPs:
  { p^#(0()) -> c_1()
  , p^#(s(x)) -> c_2()
  , minus^#(x, 0()) -> c_3()
  , minus^#(x, s(y)) -> c_4(minus^#(p(x), y)) }
Strict Trs:
  { p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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

The following argument positions are usable:
  Uargs(minus^#) = {1}, Uargs(c_4) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

            [p](x1) = [1] x1 + [2]         
                                           
                [0] = [2]                  
                                           
            [s](x1) = [1] x1 + [1]         
                                           
          [p^#](x1) = [1] x1 + [1]         
                                           
              [c_1] = [1]                  
                                           
              [c_2] = [1]                  
                                           
  [minus^#](x1, x2) = [1] x1 + [2] x2 + [2]
                                           
              [c_3] = [1]                  
                                           
          [c_4](x1) = [1] x1 + [2]         

This order satisfies following ordering constraints:

   [p(0())] = [4]        
            > [2]        
            = [0()]      
                         
  [p(s(x))] = [1] x + [3]
            > [1] x + [0]
            = [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(n^2)).

Strict DPs: { minus^#(x, s(y)) -> c_4(minus^#(p(x), y)) }
Weak DPs:
  { p^#(0()) -> c_1()
  , p^#(s(x)) -> c_2()
  , minus^#(x, 0()) -> c_3() }
Weak Trs:
  { p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

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

{ p^#(0()) -> c_1()
, p^#(s(x)) -> c_2()
, minus^#(x, 0()) -> c_3() }

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

Strict DPs: { minus^#(x, s(y)) -> c_4(minus^#(p(x), y)) }
Weak Trs:
  { p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS)' as induced by the safe mapping

 safe(p) = {1}, safe(0) = {}, safe(s) = {1}, safe(minus^#) = {1},
 safe(c_4) = {}

and precedence

 minus^# > p .

Following symbols are considered recursive:

 {p, minus^#}

The recursion depth is 2.

Further, following argument filtering is employed:

 pi(p) = 1, pi(0) = [], pi(s) = [1], pi(minus^#) = [1, 2],
 pi(c_4) = [1]

Usable defined function symbols are a subset of:

 {p, minus^#}

For your convenience, here are the satisfied ordering constraints:

  pi(minus^#(x, s(y))) =  minus^#(s(; y); x)       
                       >  c_4(minus^#(y; x);)      
                       =  pi(c_4(minus^#(p(x), y)))
                                                   
            pi(p(0())) =  0()                      
                       >= 0()                      
                       =  pi(0())                  
                                                   
           pi(p(s(x))) =  s(; x)                   
                       >  x                        
                       =  pi(x)                    
                                                   

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