/export/starexec/sandbox/solver/bin/starexec_run_ttt2 /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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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=3
   
   interpretation:
           [0]
    [a0] = [0]
           [0],
    
           [0]
    [c0] = [0]
           [1],
    
               [1 1 0]  
    [a1](x0) = [0 1 0]x0
               [0 0 0]  ,
    
               [1 0 1]     [0]
    [b1](x0) = [0 1 0]x0 + [1]
               [0 0 1]     [0],
    
               [1 0 0]     [0]
    [c1](x0) = [0 0 0]x0 + [0]
               [0 0 1]     [1],
    
           [1]
    [b0] = [1]
           [0],
    
                  [1 0 0]     [1 1 1]  
    [f](x0, x1) = [0 1 0]x0 + [1 1 1]x1
                  [1 0 1]     [1 1 1]  
   orientation:
                [1 1 1]    [1]    [1 1 0]    [0]            
    a1(b1(x)) = [0 1 0]x + [1] >= [0 1 0]x + [1] = b1(a1(x))
                [0 0 0]    [0]    [0 0 0]    [0]            
    
                [1 0 1]    [1]    [1 0 1]    [0]            
    b1(c1(x)) = [0 0 0]x + [1] >= [0 0 0]x + [0] = c1(b1(x))
                [0 0 1]    [1]    [0 0 1]    [1]            
    
                [1 1 0]    [0]    [1 0 0]             
    c1(a1(x)) = [0 0 0]x + [0] >= [0 0 0]x = a1(c1(x))
                [0 0 0]    [1]    [0 0 0]             
    
                 [1 1 1]      [1 1 0]           
    f(a0(),x1) = [1 1 1]x1 >= [0 1 0]x1 = a1(x1)
                 [1 1 1]      [0 0 0]           
    
                 [1 1 1]     [1]    [1 0 1]     [0]         
    f(b0(),x3) = [1 1 1]x3 + [1] >= [0 1 0]x3 + [1] = b1(x3)
                 [1 1 1]     [1]    [0 0 1]     [0]         
    
                 [1 1 1]     [0]    [1 0 0]     [0]         
    f(c0(),x5) = [1 1 1]x5 + [0] >= [0 0 0]x5 + [0] = c1(x5)
                 [1 1 1]     [1]    [0 0 1]     [1]         
   problem:
    c1(a1(x)) -> a1(c1(x))
    f(a0(),x1) -> a1(x1)
    f(c0(),x5) -> c1(x5)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
            [0]
     [a0] = [1]
            [1],
     
            [0]
     [c0] = [1]
            [0],
     
                [1 0 1]     [0]
     [a1](x0) = [0 0 1]x0 + [1]
                [0 1 0]     [1],
     
                [1 0 1]  
     [c1](x0) = [0 0 1]x0
                [0 1 0]  ,
     
                   [1 1 0]     [1 0 1]  
     [f](x0, x1) = [0 0 1]x0 + [0 0 1]x1
                   [0 0 1]     [0 1 0]  
    orientation:
                 [1 1 1]    [1]    [1 1 1]    [0]            
     c1(a1(x)) = [0 1 0]x + [1] >= [0 1 0]x + [1] = a1(c1(x))
                 [0 0 1]    [1]    [0 0 1]    [1]            
     
                  [1 0 1]     [1]    [1 0 1]     [0]         
     f(a0(),x1) = [0 0 1]x1 + [1] >= [0 0 1]x1 + [1] = a1(x1)
                  [0 1 0]     [1]    [0 1 0]     [1]         
     
                  [1 0 1]     [1]    [1 0 1]           
     f(c0(),x5) = [0 0 1]x5 + [0] >= [0 0 1]x5 = c1(x5)
                  [0 1 0]     [0]    [0 1 0]           
    problem:
     
    Qed