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

We add the following dependency tuples:

Strict DPs:
  { minus^#(x, 0()) -> c_1()
  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_3()
  , quot^#(s(x), s(y)) ->
    c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
    c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
        minus^#(y, s(s(z))),
        minus^#(x, s(0())))
  , plus^#(0(), y) -> c_6()
  , plus^#(s(x), y) -> c_7(plus^#(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:
  { minus^#(x, 0()) -> c_1()
  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_3()
  , quot^#(s(x), s(y)) ->
    c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
    c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
        minus^#(y, s(s(z))),
        minus^#(x, s(0())))
  , plus^#(0(), y) -> c_6()
  , plus^#(s(x), y) -> c_7(plus^#(x, y)) }
Weak Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , plus(minus(x, s(0())), minus(y, s(s(z)))) ->
    plus(minus(y, s(s(z))), minus(x, s(0())))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We estimate the number of application of {1,3,6} by applications of
Pre({1,3,6}) = {2,4,5,7}. Here rules are labeled as follows:

  DPs:
    { 1: minus^#(x, 0()) -> c_1()
    , 2: minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
    , 3: quot^#(0(), s(y)) -> c_3()
    , 4: quot^#(s(x), s(y)) ->
         c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
    , 5: plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
         c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
             minus^#(y, s(s(z))),
             minus^#(x, s(0())))
    , 6: plus^#(0(), y) -> c_6()
    , 7: plus^#(s(x), y) -> c_7(plus^#(x, y)) }

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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
    c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
        minus^#(y, s(s(z))),
        minus^#(x, s(0())))
  , plus^#(s(x), y) -> c_7(plus^#(x, y)) }
Weak DPs:
  { minus^#(x, 0()) -> c_1()
  , quot^#(0(), s(y)) -> c_3()
  , plus^#(0(), y) -> c_6() }
Weak Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , plus(minus(x, s(0())), minus(y, s(s(z)))) ->
    plus(minus(y, s(s(z))), minus(x, s(0())))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(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.

{ minus^#(x, 0()) -> c_1()
, quot^#(0(), s(y)) -> c_3()
, plus^#(0(), y) -> c_6() }

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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
    c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
        minus^#(y, s(s(z))),
        minus^#(x, s(0())))
  , plus^#(s(x), y) -> c_7(plus^#(x, y)) }
Weak Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , plus(minus(x, s(0())), minus(y, s(s(z)))) ->
    plus(minus(y, s(s(z))), minus(x, s(0())))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(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:

  { plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
    c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
        minus^#(y, s(s(z))),
        minus^#(x, s(0()))) }

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

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

We replace rewrite rules by usable rules:

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

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

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

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

DPs:
  { 3: plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
       c_3(plus^#(minus(y, s(s(z))), minus(x, s(0())))) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_4) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
      [minus](x1, x2) = [4] x1 + [0]                  
                                                      
                  [0] = [0]                           
                                                      
              [s](x1) = [1] x1 + [0]                  
                                                      
       [quot](x1, x2) = [7] x1 + [7] x2 + [0]         
                                                      
       [plus](x1, x2) = [7] x1 + [7] x2 + [0]         
                                                      
    [minus^#](x1, x2) = [0]                           
                                                      
                [c_1] = [0]                           
                                                      
            [c_2](x1) = [7] x1 + [0]                  
                                                      
     [quot^#](x1, x2) = [0]                           
                                                      
                [c_3] = [0]                           
                                                      
        [c_4](x1, x2) = [7] x1 + [7] x2 + [0]         
                                                      
     [plus^#](x1, x2) = [4]                           
                                                      
    [c_5](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [0]
                                                      
                [c_6] = [0]                           
                                                      
            [c_7](x1) = [7] x1 + [0]                  
                                                      
                  [c] = [0]                           
                                                      
            [c_1](x1) = [4] x1 + [0]                  
                                                      
        [c_2](x1, x2) = [4] x1 + [4] x2 + [0]         
                                                      
            [c_3](x1) = [1]                           
                                                      
            [c_4](x1) = [1] x1 + [0]                  
  
  The following symbols are considered usable
  
    {minus, minus^#, quot^#, plus^#}
  
  The order satisfies the following ordering constraints:
  
                                  [minus(x, 0())] =  [4] x + [0]                                       
                                                  >= [1] x + [0]                                       
                                                  =  [x]                                               
                                                                                                       
                              [minus(s(x), s(y))] =  [4] x + [0]                                       
                                                  >= [4] x + [0]                                       
                                                  =  [minus(x, y)]                                     
                                                                                                       
                            [minus^#(s(x), s(y))] =  [0]                                               
                                                  >= [0]                                               
                                                  =  [c_1(minus^#(x, y))]                              
                                                                                                       
                             [quot^#(s(x), s(y))] =  [0]                                               
                                                  >= [0]                                               
                                                  =  [c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))]   
                                                                                                       
    [plus^#(minus(x, s(0())), minus(y, s(s(z))))] =  [4]                                               
                                                  >  [1]                                               
                                                  =  [c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))]
                                                                                                       
                                [plus^#(s(x), y)] =  [4]                                               
                                                  >= [4]                                               
                                                  =  [c_4(plus^#(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^2)).

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

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

DPs:
  { 3: plus^#(s(x), y) -> c_4(plus^#(x, y)) }
Trs: { minus(s(x), s(y)) -> minus(x, y) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_4) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
      [minus](x1, x2) = [2] x1 + [0]                  
                                                      
                  [0] = [0]                           
                                                      
              [s](x1) = [1] x1 + [4]                  
                                                      
       [quot](x1, x2) = [7] x1 + [7] x2 + [0]         
                                                      
       [plus](x1, x2) = [7] x1 + [7] x2 + [0]         
                                                      
    [minus^#](x1, x2) = [0]                           
                                                      
                [c_1] = [0]                           
                                                      
            [c_2](x1) = [7] x1 + [0]                  
                                                      
     [quot^#](x1, x2) = [0]                           
                                                      
                [c_3] = [0]                           
                                                      
        [c_4](x1, x2) = [7] x1 + [7] x2 + [0]         
                                                      
     [plus^#](x1, x2) = [2] x1 + [4] x2 + [0]         
                                                      
    [c_5](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [0]
                                                      
                [c_6] = [0]                           
                                                      
            [c_7](x1) = [7] x1 + [0]                  
                                                      
                  [c] = [0]                           
                                                      
            [c_1](x1) = [1] x1 + [0]                  
                                                      
        [c_2](x1, x2) = [4] x1 + [4] x2 + [0]         
                                                      
            [c_3](x1) = [0]                           
                                                      
            [c_4](x1) = [1] x1 + [1]                  
  
  The following symbols are considered usable
  
    {minus, minus^#, quot^#, plus^#}
  
  The order satisfies the following ordering constraints:
  
                                  [minus(x, 0())] =  [2] x + [0]                                       
                                                  >= [1] x + [0]                                       
                                                  =  [x]                                               
                                                                                                       
                              [minus(s(x), s(y))] =  [2] x + [8]                                       
                                                  >  [2] x + [0]                                       
                                                  =  [minus(x, y)]                                     
                                                                                                       
                            [minus^#(s(x), s(y))] =  [0]                                               
                                                  >= [0]                                               
                                                  =  [c_1(minus^#(x, y))]                              
                                                                                                       
                             [quot^#(s(x), s(y))] =  [0]                                               
                                                  >= [0]                                               
                                                  =  [c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))]   
                                                                                                       
    [plus^#(minus(x, s(0())), minus(y, s(s(z))))] =  [4] x + [8] y + [0]                               
                                                  >= [0]                                               
                                                  =  [c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))]
                                                                                                       
                                [plus^#(s(x), y)] =  [2] x + [4] y + [8]                               
                                                  >  [2] x + [4] y + [1]                               
                                                  =  [c_4(plus^#(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^2)).

Strict DPs:
  { minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) }
Weak DPs:
  { plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
    c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
  , plus^#(s(x), y) -> c_4(plus^#(x, y)) }
Weak Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(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.

{ plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
  c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(s(x), y) -> c_4(plus^#(x, y)) }

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

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

We use the processor 'Small Polynomial Path Order (PS,2-bounded)'
to orient following rules strictly.

DPs:
  { 1: minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
  , 2: quot^#(s(x), s(y)) ->
       c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) }
Trs: { minus(s(x), s(y)) -> minus(x, y) }

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,2-bounded)' as induced by the safe mapping
  
   safe(minus) = {}, safe(0) = {}, safe(s) = {1}, safe(quot) = {},
   safe(plus) = {}, safe(minus^#) = {}, safe(c_1) = {},
   safe(c_2) = {}, safe(quot^#) = {}, safe(c_3) = {}, safe(c_4) = {},
   safe(plus^#) = {}, safe(c_5) = {}, safe(c_6) = {}, safe(c_7) = {},
   safe(c) = {}, safe(c_1) = {}, safe(c_2) = {}, safe(c_3) = {},
   safe(c_4) = {}
  
  and precedence
  
   minus > minus^#, quot^# > minus^#, minus ~ quot^# .
  
  Following symbols are considered recursive:
  
   {minus^#, quot^#}
  
  The recursion depth is 2.
  
  Further, following argument filtering is employed:
  
   pi(minus) = 1, pi(0) = [], pi(s) = [1], pi(quot) = [],
   pi(plus) = [], pi(minus^#) = [1, 2], pi(c_1) = [], pi(c_2) = [],
   pi(quot^#) = [1, 2], pi(c_3) = [], pi(c_4) = [], pi(plus^#) = [],
   pi(c_5) = [], pi(c_6) = [], pi(c_7) = [], pi(c) = [],
   pi(c_1) = [1], pi(c_2) = [1, 2], pi(c_3) = [], pi(c_4) = []
  
  Usable defined function symbols are a subset of:
  
   {minus, minus^#, quot^#, plus^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
    pi(minus^#(s(x), s(y))) =  minus^#(s(; x),  s(; y);)                        
                            >  c_1(minus^#(x,  y;);)                            
                            =  pi(c_1(minus^#(x, y)))                           
                                                                                
     pi(quot^#(s(x), s(y))) =  quot^#(s(; x),  s(; y);)                         
                            >  c_2(quot^#(x,  s(; y);),  minus^#(x,  y;);)      
                            =  pi(c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)))
                                                                                
          pi(minus(x, 0())) =  x                                                
                            >= x                                                
                            =  pi(x)                                            
                                                                                
      pi(minus(s(x), s(y))) =  s(; x)                                           
                            >  x                                                
                            =  pi(minus(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:
  { minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) }
Weak Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(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.

{ minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
, quot^#(s(x), s(y)) ->
  c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) }

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

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