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:
  { p(f(f(x))) -> q(f(g(x)))
  , p(g(g(x))) -> q(g(f(x)))
  , q(f(f(x))) -> p(f(g(x)))
  , q(g(g(x))) -> p(g(f(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:
  { p(g(g(x))) -> q(g(f(x)))
  , q(g(g(x))) -> p(g(f(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)).
  
    [p](x1) = [2 2] x1 + [0]
              [0 0]      [0]
                            
    [f](x1) = [1 0] x1 + [0]
              [0 0]      [0]
                            
    [q](x1) = [2 2] x1 + [0]
              [0 0]      [0]
                            
    [g](x1) = [1 0] x1 + [0]
              [0 1]      [1]
  
  This order satisfies the following ordering constraints:
  
    [p(f(f(x)))] =  [2 0] x + [0]
                    [0 0]     [0]
                 >= [2 0] x + [0]
                    [0 0]     [0]
                 =  [q(f(g(x)))] 
                                 
    [p(g(g(x)))] =  [2 2] x + [4]
                    [0 0]     [0]
                 >  [2 0] x + [2]
                    [0 0]     [0]
                 =  [q(g(f(x)))] 
                                 
    [q(f(f(x)))] =  [2 0] x + [0]
                    [0 0]     [0]
                 >= [2 0] x + [0]
                    [0 0]     [0]
                 =  [p(f(g(x)))] 
                                 
    [q(g(g(x)))] =  [2 2] x + [4]
                    [0 0]     [0]
                 >  [2 0] x + [2]
                    [0 0]     [0]
                 =  [p(g(f(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:
  { p(f(f(x))) -> q(f(g(x)))
  , q(f(f(x))) -> p(f(g(x))) }
Weak Trs:
  { p(g(g(x))) -> q(g(f(x)))
  , q(g(g(x))) -> p(g(f(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:
  { p(f(f(x))) -> q(f(g(x)))
  , q(f(f(x))) -> p(f(g(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)).
  
    [p](x1) = [2 1] x1 + [0]
              [0 0]      [0]
                            
    [f](x1) = [1 0] x1 + [1]
              [0 0]      [0]
                            
    [q](x1) = [2 1] x1 + [0]
              [0 0]      [0]
                            
    [g](x1) = [1 0] x1 + [0]
              [0 1]      [2]
  
  This order satisfies the following ordering constraints:
  
    [p(f(f(x)))] =  [2 0] x + [4]
                    [0 0]     [0]
                 >  [2 0] x + [2]
                    [0 0]     [0]
                 =  [q(f(g(x)))] 
                                 
    [p(g(g(x)))] =  [2 1] x + [4]
                    [0 0]     [0]
                 >= [2 0] x + [4]
                    [0 0]     [0]
                 =  [q(g(f(x)))] 
                                 
    [q(f(f(x)))] =  [2 0] x + [4]
                    [0 0]     [0]
                 >  [2 0] x + [2]
                    [0 0]     [0]
                 =  [p(f(g(x)))] 
                                 
    [q(g(g(x)))] =  [2 1] x + [4]
                    [0 0]     [0]
                 >= [2 0] x + [4]
                    [0 0]     [0]
                 =  [p(g(f(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:
  { p(f(f(x))) -> q(f(g(x)))
  , p(g(g(x))) -> q(g(f(x)))
  , q(f(f(x))) -> p(f(g(x)))
  , q(g(g(x))) -> p(g(f(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))