0.00/0.00	YES
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 zs2 zs1 ys y)
0.00/0.00	    (RULES
0.00/0.00	      split(x,nil) -> tp2(nil,nil)
0.00/0.00	      split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) == tp2(zs1,zs2), le(x,y) == true
0.00/0.00	      split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) == tp2(zs1,zs2), le(x,y) == false
0.00/0.00	      le(0,y) -> true
0.00/0.00	      le(s(x),0) -> false
0.00/0.00	      le(s(x),s(y)) -> le(x,y)
0.00/0.00	    )
0.00/0.00	    (COMMENT doi:10.2168/LMCS-8 ( 3:4 ) 2012 [64] Example 4.4 ( R_7 ) submitted by: Thomas Sternagel and Aart Middeldorp)
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 zs2 zs1 ys y)
0.00/0.00	    (RULES
0.00/0.00	      split(x,nil) -> tp2(nil,nil)
0.00/0.00	      split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) == tp2(zs1,zs2), le(x,y) == true
0.00/0.00	      split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) == tp2(zs1,zs2), le(x,y) == false
0.00/0.00	      le(0,y) -> true
0.00/0.00	      le(s(x),0) -> false
0.00/0.00	      le(s(x),s(y)) -> le(x,y)
0.00/0.00	    )
0.00/0.00	    (COMMENT doi:10.2168/LMCS-8 ( 3:4 ) 2012 [64] Example 4.4 ( R_7 ) submitted by: Thomas Sternagel and Aart Middeldorp)
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 zs2 zs1 ys y)
0.00/0.00	    (RULES
0.00/0.00	      split(x,nil) -> tp2(nil,nil)
0.00/0.00	      split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) == tp2(zs1,zs2), le(x,y) == true
0.00/0.00	      split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) == tp2(zs1,zs2), le(x,y) == false
0.00/0.00	      le(0,y) -> true
0.00/0.00	      le(s(x),0) -> false
0.00/0.00	      le(s(x),s(y)) -> le(x,y)
0.00/0.00	    )
0.00/0.00	    (COMMENT doi:10.2168/LMCS-8 ( 3:4 ) 2012 [64] Example 4.4 ( R_7 ) submitted by: Thomas Sternagel and Aart Middeldorp)
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 zs2 zs1 ys y)
0.00/0.00	    (RULES
0.00/0.00	      split(x,nil) -> tp2(nil,nil)
0.00/0.00	      split(x,cons(y,ys)) -> tp2(zs1,cons(y,zs2)) | split(x,ys) == tp2(zs1,zs2), le(x,y) == true
0.00/0.00	      split(x,cons(y,ys)) -> tp2(cons(y,zs1),zs2) | split(x,ys) == tp2(zs1,zs2), le(x,y) == false
0.00/0.00	      le(0,y) -> true
0.00/0.00	      le(s(x),0) -> false
0.00/0.00	      le(s(x),s(y)) -> le(x,y)
0.00/0.00	    )
0.00/0.00	    (COMMENT doi:10.2168/LMCS-8 ( 3:4 ) 2012 [64] Example 4.4 ( R_7 ) submitted by: Thomas Sternagel and Aart Middeldorp)
0.00/0.00	
0.00/0.00	Failed either to apply SR and U for normal 1CTRSs to the above CTRS or to prove confluence of any converted TRSs.
0.00/0.00	
0.00/0.00	Try to apply SR and U for 3DCTRSs to the above CTRS.
0.00/0.00	
0.00/0.00	Succeeded in applying U for 3DCTRSs to the above CTRS.
0.00/0.00	U(R) =
0.00/0.00	    (VAR x1 x3 x2 x5 x4)
0.00/0.00	    (RULES
0.00/0.00	      split(x1,nil) -> tp2(nil,nil)
0.00/0.00	      split(x1,cons(x2,x3)) -> u1(split(x1,x3),x1,x2,x3)
0.00/0.00	      u1(tp2(x4,x5),x1,x2,x3) -> u2(le(x1,x2),x4,x5,x1,x2,x3)
0.00/0.00	      u2(true,x4,x5,x1,x2,x3) -> tp2(x4,cons(x2,x5))
0.00/0.00	      u2(false,x4,x5,x1,x2,x3) -> tp2(cons(x2,x4),x5)
0.00/0.00	      le(0,x1) -> true
0.00/0.00	      le(s(x1),0) -> false
0.00/0.00	      le(s(x1),s(x2)) -> le(x1,x2)
0.00/0.00	    )
0.00/0.00	
0.00/0.00	U for 3DCTRSs is sound for the above CTRS.
0.00/0.00	
0.00/0.00	U(R) is confluent.
0.00/0.00	
0.00/0.00	YES
0.00/0.00	EOF