0.00/0.00	NO
0.00/0.00	
0.00/0.00	
0.00/0.00	Succeeded in reading "/export/starexec/sandbox2/benchmark/theBenchmark.trs".
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR x y z)
0.00/0.00	    (RULES
0.00/0.00	      f(y,a) -> g(x,x) | h(z) == x, y == k(z), x == b
0.00/0.00	      f(y,z) -> k(x) | z == h(a), y == k(a), a == x
0.00/0.00	      k(x) -> h(x)
0.00/0.00	      k(x) -> b
0.00/0.00	      k(x) -> a
0.00/0.00	      a -> b
0.00/0.00	      g(x,x) -> a
0.00/0.00	    )
0.00/0.00	    (COMMENT submitted by: Florian Messner used by COPS #1144)
0.00/0.00	
0.00/0.00	No "->="-rules.
0.00/0.00	
0.00/0.00	Decomposed conditions if possible.
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR x y z)
0.00/0.00	    (RULES
0.00/0.00	      f(y,a) -> g(x,x) | h(z) == x, y == k(z), x == b
0.00/0.00	      f(y,z) -> k(x) | z == h(a), y == k(a), a == x
0.00/0.00	      k(x) -> h(x)
0.00/0.00	      k(x) -> b
0.00/0.00	      k(x) -> a
0.00/0.00	      a -> b
0.00/0.00	      g(x,x) -> a
0.00/0.00	    )
0.00/0.00	    (COMMENT submitted by: Florian Messner used by COPS #1144)
0.00/0.00	
0.00/0.00	Removed infeasible rules as much as possible.
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR x y z)
0.00/0.00	    (RULES
0.00/0.00	      f(y,a) -> g(x,x) | h(z) == x, y == k(z), x == b
0.00/0.00	      f(y,z) -> k(x) | z == h(a), y == k(a), a == x
0.00/0.00	      k(x) -> h(x)
0.00/0.00	      k(x) -> b
0.00/0.00	      k(x) -> a
0.00/0.00	      a -> b
0.00/0.00	      g(x,x) -> a
0.00/0.00	    )
0.00/0.00	    (COMMENT submitted by: Florian Messner used by COPS #1144)
0.00/0.00	
0.00/0.00	Try to disprove confluence of the following (C)TRS:
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR x y z)
0.00/0.00	    (RULES
0.00/0.00	      f(y,a) -> g(x,x) | h(z) == x, y == k(z), x == b
0.00/0.00	      f(y,z) -> k(x) | z == h(a), y == k(a), a == x
0.00/0.00	      k(x) -> h(x)
0.00/0.00	      k(x) -> b
0.00/0.00	      k(x) -> a
0.00/0.00	      a -> b
0.00/0.00	      g(x,x) -> a
0.00/0.00	    )
0.00/0.00	    (COMMENT submitted by: Florian Messner used by COPS #1144)
0.00/0.00	
0.00/0.00	Succeeded in disproving confluence.
0.00/0.00	
0.00/0.00	Disproved via the following (C)TRS:
0.00/0.00	(CONDITIONTYPE ORIENTED)
0.00/0.00	(VAR x y z)
0.00/0.00	(RULES
0.00/0.00	  f(y,a) -> g(x,x) | h(z) == x, y == k(z), x == b
0.00/0.00	  f(y,z) -> k(x) | z == h(a), y == k(a), a == x
0.00/0.00	  k(x) -> h(x)
0.00/0.00	  k(x) -> b
0.00/0.00	  k(x) -> a
0.00/0.00	  a -> b
0.00/0.00	  g(x,x) -> a
0.00/0.00	)
0.00/0.00	(COMMENT submitted by: Florian Messner used by COPS #1144)
0.00/0.00	
0.00/0.00	NO
0.00/0.00	EOF