/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:
 b(b(0(),y),x) -> y
 c(c(c(y))) -> c(c(a(a(c(b(0(),y)),0()),0())))
 a(y,0()) -> b(y,0())

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                 [1 0 0]     [1 0 0]  
   [a](x0, x1) = [0 0 1]x0 + [0 0 0]x1
                 [0 0 1]     [0 0 0]  ,
   
             [1 0 1]     [0]
   [c](x0) = [0 0 0]x0 + [1]
             [0 1 0]     [0],
   
                 [1 0 0]     [1 0 0]  
   [b](x0, x1) = [0 0 1]x0 + [0 0 0]x1
                 [0 0 1]     [0 1 1]  ,
   
         [0]
   [0] = [0]
         [0]
  orientation:
                   [1 0 0]    [1 0 0]          
   b(b(0(),y),x) = [0 0 0]x + [0 1 1]y >= y = y
                   [0 1 1]    [0 1 1]          
   
                [1 1 1]    [1]    [1 1 1]    [0]                                  
   c(c(c(y))) = [0 0 0]y + [1] >= [0 0 0]y + [1] = c(c(a(a(c(b(0(),y)),0()),0())))
                [0 0 0]    [1]    [0 0 0]    [1]                                  
   
              [1 0 0]     [1 0 0]            
   a(y,0()) = [0 0 1]y >= [0 0 1]y = b(y,0())
              [0 0 1]     [0 0 1]            
  problem:
   b(b(0(),y),x) -> y
   a(y,0()) -> b(y,0())
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [a](x0, x1) = x0 + 2x1 + 2,
    
    [b](x0, x1) = x0 + 2x1 + 2,
    
    [0] = 0
   orientation:
    b(b(0(),y),x) = 2x + 2y + 4 >= y = y
    
    a(y,0()) = y + 2 >= y + 2 = b(y,0())
   problem:
    a(y,0()) -> b(y,0())
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                   [1 0 0]     [1 0 0]     [1]
     [a](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                   [0 0 0]     [0 0 0]     [0],
     
                   [1 0 0]     [1 0 0]  
     [b](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  ,
     
           [0]
     [0] = [0]
           [0]
    orientation:
                [1 0 0]    [1]    [1 0 0]            
     a(y,0()) = [0 0 0]y + [0] >= [0 0 0]y = b(y,0())
                [0 0 0]    [0]    [0 0 0]            
    problem:
     
    Qed