3.24/1.55	MAYBE
3.24/1.55	
3.24/1.55	Proof:
3.24/1.55	ConCon could not decide confluence of the system.
3.24/1.55	\cite{ALS94}, Theorem 4.1 does not apply.
3.24/1.55	This system is of type 3 or smaller.
3.24/1.55	This system is strongly deterministic.
3.24/1.55	This system is quasi-decreasing.
3.24/1.55	By \cite{A14}, Theorem 11.5.9.
3.24/1.55	This system is of type 3 or smaller.
3.24/1.55	This system is deterministic.
3.24/1.55	System R transformed to V(R) + Emb.
3.24/1.55	This system is terminating.
3.24/1.55	Call external tool:
3.24/1.55	./ttt2.sh
3.24/1.55	Input:
3.24/1.55	  f(x) -> x
3.24/1.55	  g(d, x, y) -> A
3.24/1.55	  g(d, x, y) -> y
3.24/1.55	  h(x, y) -> g(x, y, f(k))
3.24/1.55	  h(x, y) -> y
3.24/1.55	  a -> c
3.24/1.55	  a -> d
3.24/1.55	  b -> c
3.24/1.55	  b -> d
3.24/1.55	  c -> e
3.24/1.55	  c -> l
3.24/1.55	  k -> l
3.24/1.55	  k -> m
3.24/1.55	  d -> m
3.24/1.55	  h(x, y) -> x
3.24/1.55	  h(x, y) -> y
3.24/1.55	  g(x, y, z) -> x
3.24/1.55	  g(x, y, z) -> y
3.24/1.55	  g(x, y, z) -> z
3.24/1.55	  f(x) -> x
3.24/1.55	
3.24/1.55	 Polynomial Interpretation Processor:
3.24/1.55	  dimension: 1
3.24/1.55	  interpretation:
3.24/1.55	   [m] = 0,
3.24/1.55	   
3.24/1.55	   [l] = 0,
3.24/1.55	   
3.24/1.55	   [e] = 0,
3.24/1.55	   
3.24/1.55	   [b] = 7,
3.24/1.55	   
3.24/1.55	   [c] = 5,
3.24/1.55	   
3.24/1.55	   [a] = 7,
3.24/1.55	   
3.24/1.55	   [k] = 4,
3.24/1.55	   
3.24/1.55	   [h](x0, x1) = 2x0 + x1 + 2x0x0 + 7,
3.24/1.55	   
3.24/1.55	   [A] = 0,
3.24/1.55	   
3.24/1.55	   [g](x0, x1, x2) = 2x0 + x1 + x2 + 2x0x0 + 1,
3.24/1.55	   
3.24/1.55	   [d] = 6,
3.24/1.55	   
3.24/1.55	   [f](x0) = x0 + 1
3.24/1.55	  orientation:
3.24/1.55	   f(x) = x + 1 >= x = x
3.24/1.55	   
3.24/1.55	   g(d(),x,y) = x + y + 85 >= 0 = A()
3.24/1.55	   
3.24/1.55	   g(d(),x,y) = x + y + 85 >= y = y
3.24/1.55	   
3.24/1.55	   h(x,y) = 2x + 2x*x + y + 7 >= 2x + 2x*x + y + 6 = g(x,y,f(k()))
3.24/1.55	   
3.24/1.55	   h(x,y) = 2x + 2x*x + y + 7 >= y = y
3.24/1.55	   
3.24/1.55	   a() = 7 >= 5 = c()
3.24/1.55	   
3.24/1.55	   a() = 7 >= 6 = d()
3.24/1.55	   
3.24/1.55	   b() = 7 >= 5 = c()
3.24/1.55	   
3.24/1.55	   b() = 7 >= 6 = d()
3.24/1.55	   
3.24/1.55	   c() = 5 >= 0 = e()
3.24/1.55	   
3.24/1.55	   c() = 5 >= 0 = l()
3.24/1.55	   
3.24/1.55	   k() = 4 >= 0 = l()
3.24/1.55	   
3.24/1.56	   k() = 4 >= 0 = m()
3.24/1.56	   
3.24/1.56	   d() = 6 >= 0 = m()
3.24/1.56	   
3.24/1.56	   h(x,y) = 2x + 2x*x + y + 7 >= x = x
3.24/1.56	   
3.24/1.56	   g(x,y,z) = 2x + 2x*x + y + z + 1 >= x = x
3.24/1.56	   
3.24/1.56	   g(x,y,z) = 2x + 2x*x + y + z + 1 >= y = y
3.24/1.56	   
3.24/1.56	   g(x,y,z) = 2x + 2x*x + y + z + 1 >= z = z
3.24/1.56	  problem:
3.24/1.56	   
3.24/1.56	  Qed
3.24/1.56	This critical pair is not trivial.
3.24/1.56	
3.24/1.59	EOF