0.60/1.08	MAYBE
0.60/1.08	
0.60/1.08	DP problem for innermost termination.
0.60/1.08	P =
0.60/1.08	  f5#(x1) -> f4#(x1) 
0.60/1.08	  f4#(I0) -> f1#(I0) 
0.60/1.08	  f3#(I1) -> f1#(I1) 
0.60/1.08	  f1#(I2) -> f3#(-1 + I2) [1 <= I2]
0.60/1.08	  f2#(I3) -> f1#(I3) 
0.60/1.08	  f1#(I4) -> f2#(rnd1) [rnd1 = rnd1 /\ 1 <= I4]
0.60/1.08	R = 
0.60/1.08	  f5(x1) -> f4(x1) 
0.60/1.08	  f4(I0) -> f1(I0) 
0.60/1.08	  f3(I1) -> f1(I1) 
0.60/1.08	  f1(I2) -> f3(-1 + I2) [1 <= I2]
0.60/1.08	  f2(I3) -> f1(I3) 
0.60/1.08	  f1(I4) -> f2(rnd1) [rnd1 = rnd1 /\ 1 <= I4]
0.60/1.08	
0.60/1.08	The dependency graph for this problem is:
0.60/1.08	  0 -> 1
0.60/1.08	  1 -> 3, 5
0.60/1.08	  2 -> 3, 5
0.60/1.08	  3 -> 2
0.60/1.08	  4 -> 3, 5
0.60/1.08	  5 -> 4
0.60/1.08	Where:
0.60/1.08	  0) f5#(x1) -> f4#(x1) 
0.60/1.08	  1) f4#(I0) -> f1#(I0) 
0.60/1.08	  2) f3#(I1) -> f1#(I1) 
0.60/1.08	  3) f1#(I2) -> f3#(-1 + I2) [1 <= I2]
0.60/1.08	  4) f2#(I3) -> f1#(I3) 
0.60/1.08	  5) f1#(I4) -> f2#(rnd1) [rnd1 = rnd1 /\ 1 <= I4]
0.60/1.08	
0.60/1.08	We have the following SCCs.
0.60/1.08	  { 2, 3, 4, 5 }
0.60/1.08	
0.60/1.08	DP problem for innermost termination.
0.60/1.08	P =
0.60/1.08	  f3#(I1) -> f1#(I1) 
0.60/1.08	  f1#(I2) -> f3#(-1 + I2) [1 <= I2]
0.60/1.08	  f2#(I3) -> f1#(I3) 
0.60/1.08	  f1#(I4) -> f2#(rnd1) [rnd1 = rnd1 /\ 1 <= I4]
0.60/1.08	R = 
0.60/1.08	  f5(x1) -> f4(x1) 
0.60/1.08	  f4(I0) -> f1(I0) 
0.60/1.08	  f3(I1) -> f1(I1) 
0.60/1.08	  f1(I2) -> f3(-1 + I2) [1 <= I2]
0.60/1.08	  f2(I3) -> f1(I3) 
0.60/1.08	  f1(I4) -> f2(rnd1) [rnd1 = rnd1 /\ 1 <= I4]
0.60/1.08	
0.60/4.06	EOF