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

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs: { a(b(x)) -> a(c(b(x))) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
    [a](x1) = [1 2] x1 + [0]
              [0 0]      [1]
                            
    [b](x1) = [1 0] x1 + [0]
              [0 0]      [2]
                            
    [c](x1) = [1 0] x1 + [0]
              [0 0]      [0]
  
  This order satisfies the following ordering constraints:
  
    [a(b(x))] = [1 0] x + [4]
                [0 0]     [1]
              > [1 0] x + [0]
                [0 0]     [1]
              = [a(c(b(x)))] 
                             

We return to the main proof.

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

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

Empty rules are trivially bounded

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