0.60/0.68	MAYBE
0.60/0.68	
0.60/0.68	DP problem for innermost termination.
0.60/0.68	P =
0.60/0.68	  f6#(x1, x2) -> f5#(x1, x2) 
0.60/0.68	  f5#(I0, I1) -> f3#(rnd1, 1) [rnd1 = rnd1]
0.60/0.68	  f4#(I2, I3) -> f3#(I2, I3) 
0.60/0.68	  f3#(I4, I5) -> f4#(I4, I5) [1 <= I4]
0.60/0.68	  f3#(I6, I7) -> f1#(I6, 1) [I6 <= 0 /\ y1 = 0]
0.60/0.68	  f2#(I8, I9) -> f1#(I8, I9) 
0.60/0.68	  f1#(I10, I11) -> f2#(I10, I11) 
0.60/0.68	R = 
0.60/0.68	  f6(x1, x2) -> f5(x1, x2) 
0.60/0.68	  f5(I0, I1) -> f3(rnd1, 1) [rnd1 = rnd1]
0.60/0.68	  f4(I2, I3) -> f3(I2, I3) 
0.60/0.68	  f3(I4, I5) -> f4(I4, I5) [1 <= I4]
0.60/0.68	  f3(I6, I7) -> f1(I6, 1) [I6 <= 0 /\ y1 = 0]
0.60/0.68	  f2(I8, I9) -> f1(I8, I9) 
0.60/0.68	  f1(I10, I11) -> f2(I10, I11) 
0.60/0.68	
0.60/0.68	The dependency graph for this problem is:
0.60/0.68	  0 -> 1
0.60/0.68	  1 -> 3, 4
0.60/0.68	  2 -> 3, 4
0.60/0.68	  3 -> 2
0.60/0.68	  4 -> 6
0.60/0.68	  5 -> 6
0.60/0.68	  6 -> 5
0.60/0.68	Where:
0.60/0.68	  0) f6#(x1, x2) -> f5#(x1, x2) 
0.60/0.68	  1) f5#(I0, I1) -> f3#(rnd1, 1) [rnd1 = rnd1]
0.60/0.68	  2) f4#(I2, I3) -> f3#(I2, I3) 
0.60/0.68	  3) f3#(I4, I5) -> f4#(I4, I5) [1 <= I4]
0.60/0.68	  4) f3#(I6, I7) -> f1#(I6, 1) [I6 <= 0 /\ y1 = 0]
0.60/0.68	  5) f2#(I8, I9) -> f1#(I8, I9) 
0.60/0.68	  6) f1#(I10, I11) -> f2#(I10, I11) 
0.60/0.68	
0.60/0.68	We have the following SCCs.
0.60/0.68	  { 2, 3 }
0.60/0.68	  { 5, 6 }
0.60/0.68	
0.60/0.68	DP problem for innermost termination.
0.60/0.68	P =
0.60/0.68	  f2#(I8, I9) -> f1#(I8, I9) 
0.60/0.68	  f1#(I10, I11) -> f2#(I10, I11) 
0.60/0.68	R = 
0.60/0.68	  f6(x1, x2) -> f5(x1, x2) 
0.60/0.68	  f5(I0, I1) -> f3(rnd1, 1) [rnd1 = rnd1]
0.60/0.68	  f4(I2, I3) -> f3(I2, I3) 
0.60/0.68	  f3(I4, I5) -> f4(I4, I5) [1 <= I4]
0.60/0.68	  f3(I6, I7) -> f1(I6, 1) [I6 <= 0 /\ y1 = 0]
0.60/0.68	  f2(I8, I9) -> f1(I8, I9) 
0.60/0.68	  f1(I10, I11) -> f2(I10, I11) 
0.60/0.68	
0.60/3.66	EOF