1.32/1.38	MAYBE
1.32/1.38	
1.32/1.38	DP problem for innermost termination.
1.32/1.38	P =
1.32/1.38	  f5#(x1, x2) -> f1#(x1, x2) 
1.32/1.38	  f2#(I2, I3) -> f3#(I2, rnd2) [rnd2 = rnd2]
1.32/1.38	  f3#(I4, I5) -> f2#(1 + I4, I5) [1 <= I5]
1.32/1.38	  f3#(I6, I7) -> f2#(-1 + I6, I7) [I7 <= 0]
1.32/1.38	  f1#(I8, I9) -> f2#(2, I9) 
1.32/1.38	R = 
1.32/1.38	  f5(x1, x2) -> f1(x1, x2) 
1.32/1.38	  f2(I0, I1) -> f4(I0, I1) [2 <= 0]
1.32/1.38	  f2(I2, I3) -> f3(I2, rnd2) [rnd2 = rnd2]
1.32/1.38	  f3(I4, I5) -> f2(1 + I4, I5) [1 <= I5]
1.32/1.38	  f3(I6, I7) -> f2(-1 + I6, I7) [I7 <= 0]
1.32/1.38	  f1(I8, I9) -> f2(2, I9) 
1.32/1.38	
1.32/1.38	The dependency graph for this problem is:
1.32/1.38	  0 -> 4
1.32/1.38	  1 -> 2, 3
1.32/1.38	  2 -> 1
1.32/1.38	  3 -> 1
1.32/1.38	  4 -> 1
1.32/1.38	Where:
1.32/1.38	  0) f5#(x1, x2) -> f1#(x1, x2) 
1.32/1.38	  1) f2#(I2, I3) -> f3#(I2, rnd2) [rnd2 = rnd2]
1.32/1.38	  2) f3#(I4, I5) -> f2#(1 + I4, I5) [1 <= I5]
1.32/1.38	  3) f3#(I6, I7) -> f2#(-1 + I6, I7) [I7 <= 0]
1.32/1.38	  4) f1#(I8, I9) -> f2#(2, I9) 
1.32/1.38	
1.32/1.38	We have the following SCCs.
1.32/1.38	  { 1, 2, 3 }
1.32/1.38	
1.32/1.38	DP problem for innermost termination.
1.32/1.38	P =
1.32/1.38	  f2#(I2, I3) -> f3#(I2, rnd2) [rnd2 = rnd2]
1.32/1.38	  f3#(I4, I5) -> f2#(1 + I4, I5) [1 <= I5]
1.32/1.38	  f3#(I6, I7) -> f2#(-1 + I6, I7) [I7 <= 0]
1.32/1.38	R = 
1.32/1.38	  f5(x1, x2) -> f1(x1, x2) 
1.32/1.38	  f2(I0, I1) -> f4(I0, I1) [2 <= 0]
1.32/1.38	  f2(I2, I3) -> f3(I2, rnd2) [rnd2 = rnd2]
1.32/1.38	  f3(I4, I5) -> f2(1 + I4, I5) [1 <= I5]
1.32/1.38	  f3(I6, I7) -> f2(-1 + I6, I7) [I7 <= 0]
1.32/1.38	  f1(I8, I9) -> f2(2, I9) 
1.32/1.38	
1.32/4.36	EOF