/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES

Problem:
 f(a(),f(b(),x)) -> f(b(),f(a(),x))
 f(b(),f(c(),x)) -> f(c(),f(b(),x))
 f(c(),f(a(),x)) -> f(a(),f(c(),x))

Proof:
 Extended Uncurrying Processor:
  application symbol: f
  symbol table:
   c ==> c0/0 c1/1
   b ==> b0/0 b1/1
   a ==> a0/0 a1/1
   
  uncurry-rules:
   f(a0(),x1) -> a1(x1)
   f(b0(),x3) -> b1(x3)
   f(c0(),x5) -> c1(x5)
  eta-rules:
   
  problem:
   a1(b1(x)) -> b1(a1(x))
   b1(c1(x)) -> c1(b1(x))
   c1(a1(x)) -> a1(c1(x))
   f(a0(),x1) -> a1(x1)
   f(b0(),x3) -> b1(x3)
   f(c0(),x5) -> c1(x5)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c1](x0) = 2x0 + 2,
    
    [c0] = 0,
    
    [b1](x0) = 4x0 + 6,
    
    [b0] = 5,
    
    [a1](x0) = 4x0 + 6,
    
    [a0] = 3,
    
    [f](x0, x1) = x0 + 4x1 + 3
   orientation:
    a1(b1(x)) = 16x + 30 >= 16x + 30 = b1(a1(x))
    
    b1(c1(x)) = 8x + 14 >= 8x + 14 = c1(b1(x))
    
    c1(a1(x)) = 8x + 14 >= 8x + 14 = a1(c1(x))
    
    f(a0(),x1) = 4x1 + 6 >= 4x1 + 6 = a1(x1)
    
    f(b0(),x3) = 4x3 + 8 >= 4x3 + 6 = b1(x3)
    
    f(c0(),x5) = 4x5 + 3 >= 2x5 + 2 = c1(x5)
   problem:
    a1(b1(x)) -> b1(a1(x))
    b1(c1(x)) -> c1(b1(x))
    c1(a1(x)) -> a1(c1(x))
    f(a0(),x1) -> a1(x1)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                     [0]
     [c1](x0) = x0 + [1]
                     [0],
     
                [1 1 0]     [0]
     [b1](x0) = [0 1 1]x0 + [0]
                [0 0 1]     [1],
     
                [1 0 1]  
     [a1](x0) = [0 1 0]x0
                [0 0 1]  ,
     
            [0]
     [a0] = [0]
            [0],
     
                   [1 0 0]     [1 0 1]     [1]
     [f](x0, x1) = [0 0 0]x0 + [0 1 1]x1 + [0]
                   [0 0 0]     [0 0 1]     [0]
    orientation:
                 [1 1 1]    [1]    [1 1 1]    [0]            
     a1(b1(x)) = [0 1 1]x + [0] >= [0 1 1]x + [0] = b1(a1(x))
                 [0 0 1]    [1]    [0 0 1]    [1]            
     
                 [1 1 0]    [1]    [1 1 0]    [0]            
     b1(c1(x)) = [0 1 1]x + [1] >= [0 1 1]x + [1] = c1(b1(x))
                 [0 0 1]    [1]    [0 0 1]    [1]            
     
                 [1 0 1]    [0]    [1 0 1]    [0]            
     c1(a1(x)) = [0 1 0]x + [1] >= [0 1 0]x + [1] = a1(c1(x))
                 [0 0 1]    [0]    [0 0 1]    [0]            
     
                  [1 0 1]     [1]    [1 0 1]           
     f(a0(),x1) = [0 1 1]x1 + [0] >= [0 1 0]x1 = a1(x1)
                  [0 0 1]     [0]    [0 0 1]           
    problem:
     c1(a1(x)) -> a1(c1(x))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                 [1 0 1]  
      [c1](x0) = [0 0 0]x0
                 [0 0 1]  ,
      
                 [1 0 0]     [0]
      [a1](x0) = [0 0 0]x0 + [0]
                 [0 0 1]     [1]
     orientation:
                  [1 0 1]    [1]    [1 0 1]    [0]            
      c1(a1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = a1(c1(x))
                  [0 0 1]    [1]    [0 0 1]    [1]            
     problem:
      
     Qed