0.00/0.02	MAYBE
0.00/0.02	(ignored inputs)COMMENT COPS 121 150 COMMENT submitted by: Kiraku Shintani
0.00/0.02	Rewrite Rules (1):
0.00/0.02	   [ f(g(?x),g(?y)) -> f(g(?x),h(?y)),
0.00/0.02	     f(h(?x),g(?y)) -> f(g(?x),g(?y)),
0.00/0.02	     f(g(?x),h(?y)) -> f(?x,?y),
0.00/0.02	     f(h(?x),h(?y)) -> f(?y,?x),
0.00/0.02	     f(?x,?y) -> f(?y,?x),
0.00/0.02	     g(?x) -> h(?x),
0.00/0.02	     h(?x) -> g(?x) ]
0.00/0.02	Rewrite Rules (2):
0.00/0.02	   [ f(0,0) -> f(0,1),
0.00/0.02	     f(1,0) -> f(0,0),
0.00/0.02	     f(?x,?y) -> f(?y,?x) ]
0.00/0.02	both left-linear
0.00/0.02	check commutation with R1 := (1), R2 := (2)
0.00/0.02	check CPs are development closed...
0.00/0.02	...failed
0.00/0.02	check commutation with R1 := (2), R2 := (1)
0.00/0.02	check CPs are development closed...
0.00/0.02	...failed
0.00/0.02	check non-commutation
0.00/0.02	check with R1 := (1), R2 := (2)
0.00/0.02	check counter example from CPs...
0.00/0.02	check candidate: <f(0,1), f(0,0)>
0.00/0.02	check candidate: <f(0,0), f(0,1)>
0.00/0.02	check candidate: <f(g(?y_1),g(?x_1)), f(g(?x_1),h(?y_1))>
0.00/0.02	check candidate: <f(g(?y_1),h(?x_1)), f(g(?x_1),g(?y_1))>
0.00/0.02	check candidate: <f(h(?y_1),g(?x_1)), f(?x_1,?y_1)>
0.00/0.02	check candidate: <f(h(?y_1),h(?x_1)), f(?y_1,?x_1)>
0.00/0.02	check candidate: <f(?y,?x), f(?y,?x)>
0.00/0.02	...failed
0.00/0.02	check with R1 := (2), R2 := (1)
0.00/0.02	check counter example from CPs...
0.00/0.02	check candidate: <f(g(?x),h(?y)), f(g(?y),g(?x))>
0.00/0.02	check candidate: <f(g(?x),g(?y)), f(g(?y),h(?x))>
0.00/0.02	check candidate: <f(?x,?y), f(h(?y),g(?x))>
0.00/0.02	check candidate: <f(?y,?x), f(h(?y),h(?x))>
0.00/0.02	check candidate: <f(0,0), f(0,1)>
0.00/0.02	check candidate: <f(0,1), f(0,0)>
0.00/0.02	check candidate: <f(?y,?x), f(?y,?x)>
0.00/0.02	...failed
0.00/0.02	
0.00/0.02	result: Can't judge
0.00/0.02	/export/starexec/sandbox2/benchmark/theBenchmark.trs: Failure(unknown COM)
0.00/0.02	(1 msec.)
0.00/0.02	EOF