1.08/1.15	YES
1.08/1.15	
1.08/1.15	DP problem for innermost termination.
1.08/1.15	P =
1.08/1.15	  f6#(x1) -> f5#(x1) 
1.08/1.15	  f5#(I0) -> f2#(rnd1) [y1 = 5 /\ rnd1 = rnd1]
1.08/1.15	  f2#(I1) -> f3#(I1) 
1.08/1.15	  f3#(I2) -> f1#(I2) [1 + I2 <= 4]
1.08/1.15	  f1#(I4) -> f2#(1 + I4) [1 <= I4]
1.08/1.15	  f1#(I5) -> f2#(1) [I5 <= 0]
1.08/1.15	R = 
1.08/1.15	  f6(x1) -> f5(x1) 
1.08/1.15	  f5(I0) -> f2(rnd1) [y1 = 5 /\ rnd1 = rnd1]
1.08/1.15	  f2(I1) -> f3(I1) 
1.08/1.15	  f3(I2) -> f1(I2) [1 + I2 <= 4]
1.08/1.15	  f3(I3) -> f4(I3) [4 <= I3]
1.08/1.15	  f1(I4) -> f2(1 + I4) [1 <= I4]
1.08/1.15	  f1(I5) -> f2(1) [I5 <= 0]
1.08/1.15	
1.08/1.15	The dependency graph for this problem is:
1.08/1.15	  0 -> 1
1.08/1.15	  1 -> 2
1.08/1.15	  2 -> 3
1.08/1.15	  3 -> 4, 5
1.08/1.15	  4 -> 2
1.08/1.15	  5 -> 2
1.08/1.15	Where:
1.08/1.15	  0) f6#(x1) -> f5#(x1) 
1.08/1.15	  1) f5#(I0) -> f2#(rnd1) [y1 = 5 /\ rnd1 = rnd1]
1.08/1.15	  2) f2#(I1) -> f3#(I1) 
1.08/1.15	  3) f3#(I2) -> f1#(I2) [1 + I2 <= 4]
1.08/1.15	  4) f1#(I4) -> f2#(1 + I4) [1 <= I4]
1.08/1.15	  5) f1#(I5) -> f2#(1) [I5 <= 0]
1.08/1.15	
1.08/1.15	We have the following SCCs.
1.08/1.15	  { 2, 3, 4, 5 }
1.08/1.15	
1.08/1.15	DP problem for innermost termination.
1.08/1.15	P =
1.08/1.15	  f2#(I1) -> f3#(I1) 
1.08/1.15	  f3#(I2) -> f1#(I2) [1 + I2 <= 4]
1.08/1.15	  f1#(I4) -> f2#(1 + I4) [1 <= I4]
1.08/1.15	  f1#(I5) -> f2#(1) [I5 <= 0]
1.08/1.15	R = 
1.08/1.15	  f6(x1) -> f5(x1) 
1.08/1.15	  f5(I0) -> f2(rnd1) [y1 = 5 /\ rnd1 = rnd1]
1.08/1.15	  f2(I1) -> f3(I1) 
1.08/1.15	  f3(I2) -> f1(I2) [1 + I2 <= 4]
1.08/1.15	  f3(I3) -> f4(I3) [4 <= I3]
1.08/1.15	  f1(I4) -> f2(1 + I4) [1 <= I4]
1.08/1.15	  f1(I5) -> f2(1) [I5 <= 0]
1.08/1.15	
1.08/1.15	We use the extended value criterion with the projection function NU:
1.08/1.15	  NU[f1#(x0)] = -x0 + 2
1.08/1.15	  NU[f3#(x0)] = -x0 + 3
1.08/1.15	  NU[f2#(x0)] = -x0 + 3
1.08/1.15	
1.08/1.15	This gives the following inequalities:
1.08/1.15	    ==>  -I1 + 3 >= -I1 + 3
1.08/1.15	  1 + I2 <= 4  ==>  -I2 + 3 > -I2 + 2  with  -I2 + 3 >= 0
1.08/1.15	  1 <= I4  ==>  -I4 + 2 >= -(1 + I4) + 3
1.08/1.15	  I5 <= 0  ==>  -I5 + 2 >= -1 + 3
1.08/1.15	
1.08/1.15	We remove all the strictly oriented dependency pairs.
1.08/1.15	
1.08/1.15	DP problem for innermost termination.
1.08/1.15	P =
1.08/1.15	  f2#(I1) -> f3#(I1) 
1.08/1.15	  f1#(I4) -> f2#(1 + I4) [1 <= I4]
1.08/1.15	  f1#(I5) -> f2#(1) [I5 <= 0]
1.08/1.15	R = 
1.08/1.15	  f6(x1) -> f5(x1) 
1.08/1.15	  f5(I0) -> f2(rnd1) [y1 = 5 /\ rnd1 = rnd1]
1.08/1.15	  f2(I1) -> f3(I1) 
1.08/1.15	  f3(I2) -> f1(I2) [1 + I2 <= 4]
1.08/1.15	  f3(I3) -> f4(I3) [4 <= I3]
1.08/1.15	  f1(I4) -> f2(1 + I4) [1 <= I4]
1.08/1.15	  f1(I5) -> f2(1) [I5 <= 0]
1.08/1.15	
1.08/1.15	The dependency graph for this problem is:
1.08/1.15	  2 -> 
1.08/1.15	  4 -> 2
1.08/1.15	  5 -> 2
1.08/1.15	Where:
1.08/1.15	  2) f2#(I1) -> f3#(I1) 
1.08/1.15	  4) f1#(I4) -> f2#(1 + I4) [1 <= I4]
1.08/1.15	  5) f1#(I5) -> f2#(1) [I5 <= 0]
1.08/1.15	
1.08/1.15	We have the following SCCs.
1.08/1.15	  
1.08/4.12	EOF