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


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YES

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

Proof:
 Extended Uncurrying Processor:
  application symbol: f
  symbol table:
   a ==> a0/0 a1/1 a2/2
   
  uncurry-rules:
   f(a1(x1),x2) -> a2(x1,x2)
   f(a0(),x1) -> a1(x1)
  eta-rules:
   
  problem:
   a2(a1(a0()),x) -> f(x,a2(a0(),a0()))
   f(a1(x1),x2) -> a2(x1,x2)
   f(a0(),x1) -> a1(x1)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 1 0]     [1 1 0]  
    [a2](x0, x1) = [0 1 0]x0 + [1 0 0]x1
                   [0 1 0]     [0 0 1]  ,
    
               [1 1 0]     [0]
    [a1](x0) = [0 0 0]x0 + [1]
               [0 1 1]     [0],
    
           [0]
    [a0] = [0]
           [0],
    
                  [1 1 0]     [1 1 0]     [0]
    [f](x0, x1) = [1 0 0]x0 + [1 0 0]x1 + [1]
                  [0 0 1]     [0 1 1]     [1]
   orientation:
                     [1 1 0]    [1]    [1 1 0]    [0]                     
    a2(a1(a0()),x) = [1 0 0]x + [1] >= [1 0 0]x + [1] = f(x,a2(a0(),a0()))
                     [0 0 1]    [1]    [0 0 1]    [1]                     
    
                   [1 1 0]     [1 1 0]     [1]    [1 1 0]     [1 1 0]              
    f(a1(x1),x2) = [1 1 0]x1 + [1 0 0]x2 + [1] >= [0 1 0]x1 + [1 0 0]x2 = a2(x1,x2)
                   [0 1 1]     [0 1 1]     [1]    [0 1 0]     [0 0 1]              
    
                 [1 1 0]     [0]    [1 1 0]     [0]         
    f(a0(),x1) = [1 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a1(x1)
                 [0 1 1]     [1]    [0 1 1]     [0]         
   problem:
    f(a0(),x1) -> a1(x1)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                [1 0 0]  
     [a1](x0) = [0 0 0]x0
                [0 0 0]  ,
     
            [1]
     [a0] = [1]
            [0],
     
                   [1 1 0]     [1 0 0]  
     [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  
    orientation:
                  [1 0 0]     [2]    [1 0 0]           
     f(a0(),x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 = a1(x1)
                  [0 0 0]     [0]    [0 0 0]           
    problem:
     
    Qed