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

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

We add the following weak dependency pairs:

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

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)) }
Strict Trs: { a(b(x)) -> b(b(a(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)) }
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 the following constructor-restricted matrix
interpretation.

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

The order satisfies the following ordering constraints:

  [a^#(b(x))] = [1 1] x + [1]
                [0 0]     [0]
              > [1 1] x + [0]
                [0 0]     [0]
              = [c_1(a^#(x))]
                             

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(1),O(1)).

Weak DPs: { a^#(b(x)) -> c_1(a^#(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.

{ a^#(b(x)) -> c_1(a^#(x)) }

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))