0.00/0.53	MAYBE
0.00/0.53	
0.00/0.53	DP problem for innermost termination.
0.00/0.53	P =
0.00/0.53	  f4#(x1) -> f3#(x1) 
0.00/0.53	  f3#(I0) -> f1#(I0) 
0.00/0.53	  f2#(I1) -> f1#(-1 + I1) 
0.00/0.53	  f2#(I2) -> f1#(rnd1) [rnd1 = rnd1]
0.00/0.53	  f1#(I3) -> f2#(I3) [1 <= I3]
0.00/0.53	R = 
0.00/0.53	  f4(x1) -> f3(x1) 
0.00/0.53	  f3(I0) -> f1(I0) 
0.00/0.53	  f2(I1) -> f1(-1 + I1) 
0.00/0.53	  f2(I2) -> f1(rnd1) [rnd1 = rnd1]
0.00/0.53	  f1(I3) -> f2(I3) [1 <= I3]
0.00/0.53	
0.00/0.53	The dependency graph for this problem is:
0.00/0.53	  0 -> 1
0.00/0.53	  1 -> 4
0.00/0.53	  2 -> 4
0.00/0.53	  3 -> 4
0.00/0.53	  4 -> 2, 3
0.00/0.53	Where:
0.00/0.53	  0) f4#(x1) -> f3#(x1) 
0.00/0.53	  1) f3#(I0) -> f1#(I0) 
0.00/0.53	  2) f2#(I1) -> f1#(-1 + I1) 
0.00/0.53	  3) f2#(I2) -> f1#(rnd1) [rnd1 = rnd1]
0.00/0.53	  4) f1#(I3) -> f2#(I3) [1 <= I3]
0.00/0.53	
0.00/0.53	We have the following SCCs.
0.00/0.53	  { 2, 3, 4 }
0.00/0.53	
0.00/0.53	DP problem for innermost termination.
0.00/0.53	P =
0.00/0.53	  f2#(I1) -> f1#(-1 + I1) 
0.00/0.53	  f2#(I2) -> f1#(rnd1) [rnd1 = rnd1]
0.00/0.53	  f1#(I3) -> f2#(I3) [1 <= I3]
0.00/0.53	R = 
0.00/0.53	  f4(x1) -> f3(x1) 
0.00/0.53	  f3(I0) -> f1(I0) 
0.00/0.53	  f2(I1) -> f1(-1 + I1) 
0.00/0.53	  f2(I2) -> f1(rnd1) [rnd1 = rnd1]
0.00/0.53	  f1(I3) -> f2(I3) [1 <= I3]
0.00/0.53	
0.00/3.51	EOF