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

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

Strict Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add the following dependency tuples:

Strict DPs:
  { minus^#(0(), Y) -> c_1()
  , minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
  , geq^#(X, 0()) -> c_3()
  , geq^#(0(), s(Y)) -> c_4()
  , geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
  , div^#(0(), s(Y)) -> c_6()
  , div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, Y))
  , if^#(true(), X, Y) -> c_8()
  , if^#(false(), X, Y) -> c_9() }

and mark the set of starting terms.

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

Strict DPs:
  { minus^#(0(), Y) -> c_1()
  , minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
  , geq^#(X, 0()) -> c_3()
  , geq^#(0(), s(Y)) -> c_4()
  , geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
  , div^#(0(), s(Y)) -> c_6()
  , div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, Y))
  , if^#(true(), X, Y) -> c_8()
  , if^#(false(), X, Y) -> c_9() }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We estimate the number of application of {1,3,4,6,8,9} by
applications of Pre({1,3,4,6,8,9}) = {2,5,7}. Here rules are
labeled as follows:

  DPs:
    { 1: minus^#(0(), Y) -> c_1()
    , 2: minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
    , 3: geq^#(X, 0()) -> c_3()
    , 4: geq^#(0(), s(Y)) -> c_4()
    , 5: geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
    , 6: div^#(0(), s(Y)) -> c_6()
    , 7: div^#(s(X), s(Y)) ->
         c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
             geq^#(X, Y),
             div^#(minus(X, Y), s(Y)),
             minus^#(X, Y))
    , 8: if^#(true(), X, Y) -> c_8()
    , 9: if^#(false(), X, Y) -> c_9() }

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

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
  , div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, Y)) }
Weak DPs:
  { minus^#(0(), Y) -> c_1()
  , geq^#(X, 0()) -> c_3()
  , geq^#(0(), s(Y)) -> c_4()
  , div^#(0(), s(Y)) -> c_6()
  , if^#(true(), X, Y) -> c_8()
  , if^#(false(), X, Y) -> c_9() }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

{ minus^#(0(), Y) -> c_1()
, geq^#(X, 0()) -> c_3()
, geq^#(0(), s(Y)) -> c_4()
, div^#(0(), s(Y)) -> c_6()
, if^#(true(), X, Y) -> c_8()
, if^#(false(), X, Y) -> c_9() }

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

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
  , div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, Y)) }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, Y)) }

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

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
  , div^#(s(X), s(Y)) ->
    c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { minus(0(), Y) -> 0()
    , 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^1)).

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
  , div^#(s(X), s(Y)) ->
    c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We analyse the complexity of following sub-problems (R) and (S).
Problem (S) is obtained from the input problem by shifting strict
rules from (R) into the weak component:

Problem (R):
------------
  Strict DPs:
    { geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
    , div^#(s(X), s(Y)) ->
      c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
  Weak DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
  Weak Trs:
    { minus(0(), Y) -> 0()
    , minus(s(X), s(Y)) -> minus(X, Y) }
  StartTerms:
    basic terms
    Defined Symbols: minus^# geq^# div^# if^#
    Constructors: 0 s true false
  Strategy: innermost

Problem (S):
------------
  Strict DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
  Weak DPs:
    { geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
    , div^#(s(X), s(Y)) ->
      c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
  Weak Trs:
    { minus(0(), Y) -> 0()
    , minus(s(X), s(Y)) -> minus(X, Y) }
  StartTerms:
    basic terms
    Defined Symbols: minus^# geq^# div^# if^#
    Constructors: 0 s true false
  Strategy: innermost

Overall, the transformation results in the following sub-problem(s):

Generated new problems:
-----------------------
R) Strict DPs:
     { geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
     , div^#(s(X), s(Y)) ->
       c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
   Weak DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   StartTerms:
     basic terms
     Defined Symbols: minus^# geq^# div^# if^#
     Constructors: 0 s true false
   Strategy: innermost
   
   This problem was proven YES(O(1),O(n^1)).

S) Strict DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
   Weak DPs:
     { geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
     , div^#(s(X), s(Y)) ->
       c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   StartTerms:
     basic terms
     Defined Symbols: minus^# geq^# div^# if^#
     Constructors: 0 s true false
   Strategy: innermost
   
   This problem was proven YES(O(1),O(n^1)).


Proofs for generated problems:
------------------------------
R) We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs:
     { geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
     , div^#(s(X), s(Y)) ->
       c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
   Weak DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(n^1))
   
   Due to missing edges in the dependency-graph, the right-hand sides
   of following rules could be simplified:
   
     { div^#(s(X), s(Y)) ->
       c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
   
   We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs:
     { geq^#(s(X), s(Y)) -> c_1(geq^#(X, Y))
     , div^#(s(X), s(Y)) -> c_2(geq^#(X, Y), minus^#(X, Y)) }
   Weak DPs: { minus^#(s(X), s(Y)) -> c_3(minus^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(n^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_3(minus^#(X, Y)) }
   
   We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs:
     { geq^#(s(X), s(Y)) -> c_1(geq^#(X, Y))
     , div^#(s(X), s(Y)) -> c_2(geq^#(X, Y), minus^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(n^1))
   
   Due to missing edges in the dependency-graph, the right-hand sides
   of following rules could be simplified:
   
     { div^#(s(X), s(Y)) -> c_2(geq^#(X, Y), minus^#(X, Y)) }
   
   We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs:
     { geq^#(s(X), s(Y)) -> c_1(geq^#(X, Y))
     , div^#(s(X), s(Y)) -> c_2(geq^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(n^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(n^1)).
   
   Strict DPs:
     { geq^#(s(X), s(Y)) -> c_1(geq^#(X, Y))
     , div^#(s(X), s(Y)) -> c_2(geq^#(X, Y)) }
   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: geq^#(s(X), s(Y)) -> c_1(geq^#(X, Y))
     , 2: div^#(s(X), s(Y)) -> c_2(geq^#(X, Y)) }
   
   Sub-proof:
   ----------
     The input was oriented with the instance of 'Small Polynomial Path
     Order (PS,1-bounded)' as induced by the safe mapping
     
      safe(minus) = {}, safe(0) = {}, safe(s) = {1}, safe(geq) = {},
      safe(true) = {}, safe(false) = {}, safe(div) = {}, safe(if) = {},
      safe(minus^#) = {}, safe(c_1) = {}, safe(c_2) = {},
      safe(geq^#) = {1}, safe(c_3) = {}, safe(c_4) = {}, safe(c_5) = {},
      safe(div^#) = {}, safe(c_6) = {}, safe(c_7) = {}, safe(if^#) = {},
      safe(c_8) = {}, safe(c_9) = {}, safe(c) = {}, safe(c_1) = {},
      safe(c_2) = {}, safe(c_3) = {}, safe(c) = {}, safe(c_1) = {},
      safe(c_2) = {}, safe(c_3) = {}, safe(c) = {}, safe(c_1) = {},
      safe(c_2) = {}
     
     and precedence
     
      div^# > geq^# .
     
     Following symbols are considered recursive:
     
      {geq^#}
     
     The recursion depth is 1.
     
     Further, following argument filtering is employed:
     
      pi(minus) = [], pi(0) = [], pi(s) = [1], pi(geq) = [],
      pi(true) = [], pi(false) = [], pi(div) = [], pi(if) = [],
      pi(minus^#) = [], pi(c_1) = [], pi(c_2) = [], pi(geq^#) = [2],
      pi(c_3) = [], pi(c_4) = [], pi(c_5) = [], pi(div^#) = [2],
      pi(c_6) = [], pi(c_7) = [], pi(if^#) = [], pi(c_8) = [],
      pi(c_9) = [], pi(c) = [], pi(c_1) = [], pi(c_2) = [], pi(c_3) = [],
      pi(c) = [], pi(c_1) = [], pi(c_2) = [], pi(c_3) = [], pi(c) = [],
      pi(c_1) = [1], pi(c_2) = [1]
     
     Usable defined function symbols are a subset of:
     
      {minus^#, geq^#, div^#, if^#}
     
     For your convenience, here are the satisfied ordering constraints:
     
       pi(geq^#(s(X), s(Y))) = geq^#(s(; Y);)      
                             > c_1(geq^#(Y;);)     
                             = pi(c_1(geq^#(X, Y)))
                                                   
       pi(div^#(s(X), s(Y))) = div^#(s(; Y);)      
                             > c_2(geq^#(Y;);)     
                             = pi(c_2(geq^#(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:
     { geq^#(s(X), s(Y)) -> c_1(geq^#(X, Y))
     , div^#(s(X), s(Y)) -> c_2(geq^#(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.
   
   { geq^#(s(X), s(Y)) -> c_1(geq^#(X, Y))
   , div^#(s(X), s(Y)) -> c_2(geq^#(X, Y)) }
   
   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

S) We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
   Weak DPs:
     { geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
     , div^#(s(X), s(Y)) ->
       c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(n^1))
   
   Due to missing edges in the dependency-graph, the right-hand sides
   of following rules could be simplified:
   
     { div^#(s(X), s(Y)) ->
       c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
   
   We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
   Weak DPs:
     { geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
     , div^#(s(X), s(Y)) -> c_3(geq^#(X, Y), minus^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(n^1))
   
   The following weak DPs constitute a sub-graph of the DG that is
   closed under successors. The DPs are removed.
   
   { geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y)) }
   
   We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
   Weak DPs: { div^#(s(X), s(Y)) -> c_3(geq^#(X, Y), minus^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(n^1))
   
   Due to missing edges in the dependency-graph, the right-hand sides
   of following rules could be simplified:
   
     { div^#(s(X), s(Y)) -> c_3(geq^#(X, Y), minus^#(X, Y)) }
   
   We are left with following problem, upon which TcT provides the
   certificate YES(O(1),O(n^1)).
   
   Strict DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
   Weak DPs: { div^#(s(X), s(Y)) -> c_2(minus^#(X, Y)) }
   Weak Trs:
     { minus(0(), Y) -> 0()
     , minus(s(X), s(Y)) -> minus(X, Y) }
   Obligation:
     innermost runtime complexity
   Answer:
     YES(O(1),O(n^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(n^1)).
   
   Strict DPs: { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y)) }
   Weak DPs: { div^#(s(X), s(Y)) -> c_2(minus^#(X, Y)) }
   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: minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y))
     , 2: div^#(s(X), s(Y)) -> c_2(minus^#(X, Y)) }
   
   Sub-proof:
   ----------
     The input was oriented with the instance of 'Small Polynomial Path
     Order (PS,1-bounded)' as induced by the safe mapping
     
      safe(minus) = {}, safe(0) = {}, safe(s) = {1}, safe(geq) = {},
      safe(true) = {}, safe(false) = {}, safe(div) = {}, safe(if) = {},
      safe(minus^#) = {1}, safe(c_1) = {}, safe(c_2) = {},
      safe(geq^#) = {}, safe(c_3) = {}, safe(c_4) = {}, safe(c_5) = {},
      safe(div^#) = {}, safe(c_6) = {}, safe(c_7) = {}, safe(if^#) = {},
      safe(c_8) = {}, safe(c_9) = {}, safe(c) = {}, safe(c_1) = {},
      safe(c_2) = {}, safe(c_3) = {}, safe(c) = {}, safe(c_1) = {},
      safe(c_2) = {}, safe(c_3) = {}, safe(c) = {}, safe(c_1) = {},
      safe(c_2) = {}
     
     and precedence
     
      div^# > minus^# .
     
     Following symbols are considered recursive:
     
      {minus^#}
     
     The recursion depth is 1.
     
     Further, following argument filtering is employed:
     
      pi(minus) = [], pi(0) = [], pi(s) = [1], pi(geq) = [],
      pi(true) = [], pi(false) = [], pi(div) = [], pi(if) = [],
      pi(minus^#) = [2], pi(c_1) = [], pi(c_2) = [], pi(geq^#) = [],
      pi(c_3) = [], pi(c_4) = [], pi(c_5) = [], pi(div^#) = [2],
      pi(c_6) = [], pi(c_7) = [], pi(if^#) = [], pi(c_8) = [],
      pi(c_9) = [], pi(c) = [], pi(c_1) = [], pi(c_2) = [], pi(c_3) = [],
      pi(c) = [], pi(c_1) = [], pi(c_2) = [], pi(c_3) = [], pi(c) = [],
      pi(c_1) = [1], pi(c_2) = [1]
     
     Usable defined function symbols are a subset of:
     
      {minus^#, geq^#, div^#, if^#}
     
     For your convenience, here are the satisfied ordering constraints:
     
       pi(minus^#(s(X), s(Y))) = minus^#(s(; Y);)      
                               > c_1(minus^#(Y;);)     
                               = pi(c_1(minus^#(X, Y)))
                                                       
         pi(div^#(s(X), s(Y))) = div^#(s(; Y);)        
                               > c_2(minus^#(Y;);)     
                               = pi(c_2(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))
     , div^#(s(X), s(Y)) -> c_2(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))
   , div^#(s(X), s(Y)) -> c_2(minus^#(X, Y)) }
   
   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^1))