1.37/1.41	MAYBE
1.37/1.41	
1.37/1.41	DP problem for innermost termination.
1.37/1.41	P =
1.37/1.41	  f4#(x1, x2) -> f3#(x1, x2) 
1.37/1.41	  f3#(I0, I1) -> f1#(I0, I1) 
1.37/1.41	  f2#(I2, I3) -> f1#(-1 + I2, I3) 
1.37/1.41	  f2#(I4, I5) -> f1#(rnd1, -1 + I5) [rnd1 = rnd1]
1.37/1.41	  f1#(I6, I7) -> f2#(I6, I7) [1 <= I6]
1.37/1.41	R = 
1.37/1.41	  f4(x1, x2) -> f3(x1, x2) 
1.37/1.41	  f3(I0, I1) -> f1(I0, I1) 
1.37/1.41	  f2(I2, I3) -> f1(-1 + I2, I3) 
1.37/1.41	  f2(I4, I5) -> f1(rnd1, -1 + I5) [rnd1 = rnd1]
1.37/1.41	  f1(I6, I7) -> f2(I6, I7) [1 <= I6]
1.37/1.41	
1.37/1.41	The dependency graph for this problem is:
1.37/1.41	  0 -> 1
1.37/1.41	  1 -> 4
1.37/1.41	  2 -> 4
1.37/1.41	  3 -> 4
1.37/1.41	  4 -> 2, 3
1.37/1.41	Where:
1.37/1.41	  0) f4#(x1, x2) -> f3#(x1, x2) 
1.37/1.41	  1) f3#(I0, I1) -> f1#(I0, I1) 
1.37/1.41	  2) f2#(I2, I3) -> f1#(-1 + I2, I3) 
1.37/1.41	  3) f2#(I4, I5) -> f1#(rnd1, -1 + I5) [rnd1 = rnd1]
1.37/1.41	  4) f1#(I6, I7) -> f2#(I6, I7) [1 <= I6]
1.37/1.41	
1.37/1.41	We have the following SCCs.
1.37/1.41	  { 2, 3, 4 }
1.37/1.41	
1.37/1.41	DP problem for innermost termination.
1.37/1.41	P =
1.37/1.41	  f2#(I2, I3) -> f1#(-1 + I2, I3) 
1.37/1.41	  f2#(I4, I5) -> f1#(rnd1, -1 + I5) [rnd1 = rnd1]
1.37/1.41	  f1#(I6, I7) -> f2#(I6, I7) [1 <= I6]
1.37/1.41	R = 
1.37/1.41	  f4(x1, x2) -> f3(x1, x2) 
1.37/1.41	  f3(I0, I1) -> f1(I0, I1) 
1.37/1.41	  f2(I2, I3) -> f1(-1 + I2, I3) 
1.37/1.41	  f2(I4, I5) -> f1(rnd1, -1 + I5) [rnd1 = rnd1]
1.37/1.41	  f1(I6, I7) -> f2(I6, I7) [1 <= I6]
1.37/1.41	
1.37/4.39	EOF