YES(?,O(n^1))

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

Strict Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(b(), X, c()) -> a__f(X, a__c(), X)
  , a__c() -> b()
  , a__c() -> c()
  , mark(b()) -> b()
  , mark(c()) -> a__c()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We add following dependency tuples:

Strict DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__f^#(b(), X, c()) -> c_2(a__f^#(X, a__c(), X), a__c^#())
  , a__c^#() -> c_3()
  , a__c^#() -> c_4()
  , mark^#(b()) -> c_5()
  , mark^#(c()) -> c_6(a__c^#())
  , mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }

and mark the set of starting terms.

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

Strict DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__f^#(b(), X, c()) -> c_2(a__f^#(X, a__c(), X), a__c^#())
  , a__c^#() -> c_3()
  , a__c^#() -> c_4()
  , mark^#(b()) -> c_5()
  , mark^#(c()) -> c_6(a__c^#())
  , mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(b(), X, c()) -> a__f(X, a__c(), X)
  , a__c() -> b()
  , a__c() -> c()
  , mark(b()) -> b()
  , mark(c()) -> a__c()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

  DPs:
    { 1: a__f^#(X1, X2, X3) -> c_1()
    , 2: a__f^#(b(), X, c()) -> c_2(a__f^#(X, a__c(), X), a__c^#())
    , 3: a__c^#() -> c_3()
    , 4: a__c^#() -> c_4()
    , 5: mark^#(b()) -> c_5()
    , 6: mark^#(c()) -> c_6(a__c^#())
    , 7: mark^#(f(X1, X2, X3)) ->
         c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }

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

Strict DPs:
  { a__f^#(b(), X, c()) -> c_2(a__f^#(X, a__c(), X), a__c^#())
  , mark^#(c()) -> c_6(a__c^#())
  , mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }
Weak DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__c^#() -> c_3()
  , a__c^#() -> c_4()
  , mark^#(b()) -> c_5() }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(b(), X, c()) -> a__f(X, a__c(), X)
  , a__c() -> b()
  , a__c() -> c()
  , mark(b()) -> b()
  , mark(c()) -> a__c()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

  DPs:
    { 1: a__f^#(b(), X, c()) -> c_2(a__f^#(X, a__c(), X), a__c^#())
    , 2: mark^#(c()) -> c_6(a__c^#())
    , 3: mark^#(f(X1, X2, X3)) ->
         c_7(a__f^#(X1, mark(X2), X3), mark^#(X2))
    , 4: a__f^#(X1, X2, X3) -> c_1()
    , 5: a__c^#() -> c_3()
    , 6: a__c^#() -> c_4()
    , 7: mark^#(b()) -> c_5() }

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

Strict DPs:
  { mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }
Weak DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__f^#(b(), X, c()) -> c_2(a__f^#(X, a__c(), X), a__c^#())
  , a__c^#() -> c_3()
  , a__c^#() -> c_4()
  , mark^#(b()) -> c_5()
  , mark^#(c()) -> c_6(a__c^#()) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(b(), X, c()) -> a__f(X, a__c(), X)
  , a__c() -> b()
  , a__c() -> c()
  , mark(b()) -> b()
  , mark(c()) -> a__c()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

{ a__f^#(X1, X2, X3) -> c_1()
, a__f^#(b(), X, c()) -> c_2(a__f^#(X, a__c(), X), a__c^#())
, a__c^#() -> c_3()
, a__c^#() -> c_4()
, mark^#(b()) -> c_5()
, mark^#(c()) -> c_6(a__c^#()) }

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

Strict DPs:
  { mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(b(), X, c()) -> a__f(X, a__c(), X)
  , a__c() -> b()
  , a__c() -> c()
  , mark(b()) -> b()
  , mark(c()) -> a__c()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

  { mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }

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

Strict DPs: { mark^#(f(X1, X2, X3)) -> c_1(mark^#(X2)) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(b(), X, c()) -> a__f(X, a__c(), X)
  , a__c() -> b()
  , a__c() -> c()
  , mark(b()) -> b()
  , mark(c()) -> a__c()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,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(n^1)).

Strict DPs: { mark^#(f(X1, X2, X3)) -> c_1(mark^#(X2)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

 safe(f) = {1, 2, 3}, safe(mark^#) = {}, safe(c_1) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {mark^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(f) = [1, 2, 3], pi(mark^#) = [1], pi(c_1) = [1]

Usable defined function symbols are a subset of:

 {mark^#}

For your convenience, here are the satisfied ordering constraints:

  pi(mark^#(f(X1, X2, X3))) = mark^#(f(; X1,  X2,  X3);)
                            > c_1(mark^#(X2;);)         
                            = pi(c_1(mark^#(X2)))       
                                                        

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