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:
  { and(not(not(x)), y, not(z)) -> and(y, band(x, z), 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: { and(not(not(x)), y, not(z)) -> and(y, band(x, z), 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)).
  
    [and](x1, x2, x3) = [1 3] x1 + [1 3] x2 + [3 0] x3 + [1]
                        [2 3]      [2 3]      [2 0]      [1]
                                                            
            [not](x1) = [1 0] x1 + [0]                      
                        [1 0]      [1]                      
                                                            
       [band](x1, x2) = [1 0] x1 + [1 0] x2 + [1]           
                        [0 0]      [0 0]      [0]           
  
  This order satisfies the following ordering constraints:
  
    [and(not(not(x)), y, not(z))] = [4 0] x + [1 3] y + [3 0] z + [4]
                                    [5 0]     [2 3]     [2 0]     [4]
                                  > [4 0] x + [1 3] y + [1 0] z + [2]
                                    [4 0]     [2 3]     [2 0]     [3]
                                  = [and(y, band(x, z), 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: { and(not(not(x)), y, not(z)) -> and(y, band(x, z), 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))