/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

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

Proof:
 Extended Uncurrying Processor:
  application symbol: f
  symbol table:
   b ==> b0/0 b1/1
   a ==> a0/0 a1/1
   
  uncurry-rules:
   f(a0(),x1) -> a1(x1)
   f(b0(),x3) -> b1(x3)
  eta-rules:
   
  problem:
   a1(b1(a1(x))) -> a1(b1(b1(a1(x))))
   b1(b1(b1(x))) -> b1(b1(x))
   f(a0(),x1) -> a1(x1)
   f(b0(),x3) -> b1(x3)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [b1](x0) = x0,
    
    [b0] = 2,
    
    [a1](x0) = x0,
    
    [a0] = 1,
    
    [f](x0, x1) = 5x0 + 4x1 + 2
   orientation:
    a1(b1(a1(x))) = x >= x = a1(b1(b1(a1(x))))
    
    b1(b1(b1(x))) = x >= x = b1(b1(x))
    
    f(a0(),x1) = 4x1 + 7 >= x1 = a1(x1)
    
    f(b0(),x3) = 4x3 + 12 >= x3 = b1(x3)
   problem:
    a1(b1(a1(x))) -> a1(b1(b1(a1(x))))
    b1(b1(b1(x))) -> b1(b1(x))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                [1 0 0]  
     [b1](x0) = [0 0 0]x0
                [0 1 0]  ,
     
                [1 0 1]     [0]
     [a1](x0) = [0 1 0]x0 + [1]
                [0 0 0]     [0]
    orientation:
                     [1 1 1]    [1]    [1 0 1]    [0]                    
     a1(b1(a1(x))) = [0 0 0]x + [1] >= [0 0 0]x + [1] = a1(b1(b1(a1(x))))
                     [0 0 0]    [0]    [0 0 0]    [0]                    
     
                     [1 0 0]     [1 0 0]             
     b1(b1(b1(x))) = [0 0 0]x >= [0 0 0]x = b1(b1(x))
                     [0 0 0]     [0 0 0]             
    problem:
     b1(b1(b1(x))) -> b1(b1(x))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                 [1 1 0]     [0]
      [b1](x0) = [0 0 1]x0 + [0]
                 [0 0 0]     [1]
     orientation:
                      [1 1 1]    [1]    [1 1 1]    [0]            
      b1(b1(b1(x))) = [0 0 0]x + [1] >= [0 0 0]x + [1] = b1(b1(x))
                      [0 0 0]    [1]    [0 0 0]    [1]            
     problem:
      
     Qed