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:
  { a(b(x)) -> b(a(x))
  , a(c(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { a^#(b(x)) -> c_1(a^#(x))
  , a^#(c(x)) -> c_2() }

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:
  { a^#(b(x)) -> c_1(a^#(x))
  , a^#(c(x)) -> c_2() }
Strict Trs:
  { a(b(x)) -> b(a(x))
  , a(c(x)) -> x }
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:
  { a^#(b(x)) -> c_1(a^#(x))
  , a^#(c(x)) -> c_2() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

The following argument positions are usable:
  Uargs(c_1) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

    [b](x1) = [1] x1 + [2]
                          
    [c](x1) = [2]         
                          
  [a^#](x1) = [1]         
                          
  [c_1](x1) = [1] x1 + [0]
                          
      [c_2] = [0]         

This order satisfies following ordering constraints:


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^1)).

Strict DPs: { a^#(b(x)) -> c_1(a^#(x)) }
Weak DPs: { a^#(c(x)) -> c_2() }
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^#(c(x)) -> c_2() }

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

Strict DPs: { a^#(b(x)) -> c_1(a^#(x)) }
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(b) = {1}, safe(a^#) = {}, safe(c_1) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {a^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(b) = [1], pi(a^#) = [1], pi(c_1) = [1]

Usable defined function symbols are a subset of:

 {a^#}

For your convenience, here are the satisfied ordering constraints:

  pi(a^#(b(x))) = a^#(b(; x);)   
                > c_1(a^#(x;);)  
                = pi(c_1(a^#(x)))
                                 

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