1.51/1.53	MAYBE
1.51/1.53	
1.51/1.53	DP problem for innermost termination.
1.51/1.53	P =
1.51/1.53	  f6#(x1, x2) -> f5#(x1, x2) 
1.51/1.53	  f5#(I0, I1) -> f2#(0, 99) 
1.51/1.53	  f2#(I2, I3) -> f3#(I2, I3) 
1.51/1.53	  f3#(I4, I5) -> f1#(I4, I5) [1 + I4 <= I5]
1.51/1.53	  f1#(I8, I9) -> f2#(I8, rnd2) [rnd2 = rnd2]
1.51/1.53	  f1#(I10, I11) -> f2#(rnd1, I11) [rnd1 = rnd1]
1.51/1.53	R = 
1.51/1.53	  f6(x1, x2) -> f5(x1, x2) 
1.51/1.53	  f5(I0, I1) -> f2(0, 99) 
1.51/1.53	  f2(I2, I3) -> f3(I2, I3) 
1.51/1.53	  f3(I4, I5) -> f1(I4, I5) [1 + I4 <= I5]
1.51/1.53	  f3(I6, I7) -> f4(I6, I7) [I7 <= I6]
1.51/1.53	  f1(I8, I9) -> f2(I8, rnd2) [rnd2 = rnd2]
1.51/1.53	  f1(I10, I11) -> f2(rnd1, I11) [rnd1 = rnd1]
1.51/1.53	
1.51/1.53	The dependency graph for this problem is:
1.51/1.53	  0 -> 1
1.51/1.53	  1 -> 2
1.51/1.53	  2 -> 3
1.51/1.53	  3 -> 4, 5
1.51/1.53	  4 -> 2
1.51/1.53	  5 -> 2
1.51/1.53	Where:
1.51/1.53	  0) f6#(x1, x2) -> f5#(x1, x2) 
1.51/1.53	  1) f5#(I0, I1) -> f2#(0, 99) 
1.51/1.53	  2) f2#(I2, I3) -> f3#(I2, I3) 
1.51/1.53	  3) f3#(I4, I5) -> f1#(I4, I5) [1 + I4 <= I5]
1.51/1.53	  4) f1#(I8, I9) -> f2#(I8, rnd2) [rnd2 = rnd2]
1.51/1.53	  5) f1#(I10, I11) -> f2#(rnd1, I11) [rnd1 = rnd1]
1.51/1.53	
1.51/1.53	We have the following SCCs.
1.51/1.53	  { 2, 3, 4, 5 }
1.51/1.53	
1.51/1.53	DP problem for innermost termination.
1.51/1.53	P =
1.51/1.53	  f2#(I2, I3) -> f3#(I2, I3) 
1.51/1.53	  f3#(I4, I5) -> f1#(I4, I5) [1 + I4 <= I5]
1.51/1.53	  f1#(I8, I9) -> f2#(I8, rnd2) [rnd2 = rnd2]
1.51/1.53	  f1#(I10, I11) -> f2#(rnd1, I11) [rnd1 = rnd1]
1.51/1.53	R = 
1.51/1.53	  f6(x1, x2) -> f5(x1, x2) 
1.51/1.53	  f5(I0, I1) -> f2(0, 99) 
1.51/1.53	  f2(I2, I3) -> f3(I2, I3) 
1.51/1.53	  f3(I4, I5) -> f1(I4, I5) [1 + I4 <= I5]
1.51/1.53	  f3(I6, I7) -> f4(I6, I7) [I7 <= I6]
1.51/1.53	  f1(I8, I9) -> f2(I8, rnd2) [rnd2 = rnd2]
1.51/1.53	  f1(I10, I11) -> f2(rnd1, I11) [rnd1 = rnd1]
1.51/1.53	
1.51/4.51	EOF