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

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { d(u(x)) -> c(x)
  , c(u(x)) -> b(x)
  , b(u(x)) -> a(e(x))
  , v(e(x)) -> 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).
  
    [d](x1) = [1] x1 + [3]
                          
    [e](x1) = [1] x1 + [0]
                          
    [u](x1) = [1] x1 + [3]
                          
    [c](x1) = [1] x1 + [3]
                          
    [b](x1) = [1] x1 + [3]
                          
    [v](x1) = [3] x1 + [3]
                          
    [a](x1) = [1] x1 + [1]
  
  This order satisfies the following ordering constraints:
  
       [d(x)] =  [1] x + [3]
              >= [1] x + [3]
              =  [e(u(x))]  
                            
    [d(u(x))] =  [1] x + [6]
              >  [1] x + [3]
              =  [c(x)]     
                            
    [c(u(x))] =  [1] x + [6]
              >  [1] x + [3]
              =  [b(x)]     
                            
    [b(u(x))] =  [1] x + [6]
              >  [1] x + [1]
              =  [a(e(x))]  
                            
    [v(e(x))] =  [3] x + [3]
              >  [1] x + [0]
              =  [x]        
                            

We return to the main proof.

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

Strict Trs: { d(x) -> e(u(x)) }
Weak Trs:
  { d(u(x)) -> c(x)
  , c(u(x)) -> b(x)
  , b(u(x)) -> a(e(x))
  , v(e(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs: { d(x) -> e(u(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).
  
    [d](x1) = [3] x1 + [3]
                          
    [e](x1) = [1] x1 + [1]
                          
    [u](x1) = [1] x1 + [1]
                          
    [c](x1) = [3] x1 + [3]
                          
    [b](x1) = [3] x1 + [3]
                          
    [v](x1) = [3] x1 + [3]
                          
    [a](x1) = [1] x1 + [3]
  
  This order satisfies the following ordering constraints:
  
       [d(x)] = [3] x + [3]
              > [1] x + [2]
              = [e(u(x))]  
                           
    [d(u(x))] = [3] x + [6]
              > [3] x + [3]
              = [c(x)]     
                           
    [c(u(x))] = [3] x + [6]
              > [3] x + [3]
              = [b(x)]     
                           
    [b(u(x))] = [3] x + [6]
              > [1] x + [4]
              = [a(e(x))]  
                           
    [v(e(x))] = [3] x + [6]
              > [1] x + [0]
              = [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:
  { d(x) -> e(u(x))
  , d(u(x)) -> c(x)
  , c(u(x)) -> b(x)
  , b(u(x)) -> a(e(x))
  , v(e(x)) -> 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))