/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:
 .(1(),x) -> x
 .(x,1()) -> x
 .(i(x),x) -> 1()
 .(x,i(x)) -> 1()
 i(1()) -> 1()
 i(i(x)) -> x
 .(i(y),.(y,z)) -> z
 .(y,.(i(y),z)) -> z

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [i](x0) = x0 + 4,
   
   [.](x0, x1) = x0 + x1 + 7,
   
   [1] = 0
  orientation:
   .(1(),x) = x + 7 >= x = x
   
   .(x,1()) = x + 7 >= x = x
   
   .(i(x),x) = 2x + 11 >= 0 = 1()
   
   .(x,i(x)) = 2x + 11 >= 0 = 1()
   
   i(1()) = 4 >= 0 = 1()
   
   i(i(x)) = x + 8 >= x = x
   
   .(i(y),.(y,z)) = 2y + z + 18 >= z = z
   
   .(y,.(i(y),z)) = 2y + z + 18 >= z = z
  problem:
   
  Qed