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

We add the following dependency tuples:

Strict DPs:
  { pred^#(s(x)) -> c_1()
  , minus^#(x, s(y)) -> c_2(pred^#(minus(x, y)), minus^#(x, y))
  , minus^#(x, 0()) -> c_3()
  , quot^#(s(x), s(y)) ->
    c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , quot^#(0(), s(y)) -> c_5() }

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

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

  DPs:
    { 1: pred^#(s(x)) -> c_1()
    , 2: minus^#(x, s(y)) -> c_2(pred^#(minus(x, y)), minus^#(x, y))
    , 3: minus^#(x, 0()) -> c_3()
    , 4: quot^#(s(x), s(y)) ->
         c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
    , 5: quot^#(0(), s(y)) -> c_5() }

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

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

{ pred^#(s(x)) -> c_1()
, minus^#(x, 0()) -> c_3()
, quot^#(0(), s(y)) -> c_5() }

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

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

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

We replace rewrite rules by usable rules:

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

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

Strict DPs:
  { minus^#(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:
  { pred(s(x)) -> x
  , minus(x, s(y)) -> pred(minus(x, y))
  , minus(x, 0()) -> x }
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^#(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: { pred(s(x)) -> x }

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