0.00/0.49	YES
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  f7#(x1) -> f6#(x1) 
0.00/0.49	  f6#(I0) -> f5#(rnd1) [y1 = 0 /\ rnd1 = rnd1]
0.00/0.49	  f4#(I1) -> f3#(I1) 
0.00/0.49	  f5#(I2) -> f4#(I2) [1 <= I2]
0.00/0.49	  f5#(I3) -> f1#(I3) [I3 <= 0]
0.00/0.49	  f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20]
0.00/0.49	  f3#(I5) -> f1#(I5) [20 <= I5]
0.00/0.49	R = 
0.00/0.49	  f7(x1) -> f6(x1) 
0.00/0.49	  f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1]
0.00/0.49	  f4(I1) -> f3(I1) 
0.00/0.49	  f5(I2) -> f4(I2) [1 <= I2]
0.00/0.49	  f5(I3) -> f1(I3) [I3 <= 0]
0.00/0.49	  f3(I4) -> f4(1 + I4) [1 + I4 <= 20]
0.00/0.49	  f3(I5) -> f1(I5) [20 <= I5]
0.00/0.49	  f1(I6) -> f2(I6) 
0.00/0.49	
0.00/0.49	The dependency graph for this problem is:
0.00/0.49	  0 -> 1
0.00/0.49	  1 -> 3, 4
0.00/0.49	  2 -> 5, 6
0.00/0.49	  3 -> 2
0.00/0.49	  4 -> 
0.00/0.49	  5 -> 2
0.00/0.49	  6 -> 
0.00/0.49	Where:
0.00/0.49	  0) f7#(x1) -> f6#(x1) 
0.00/0.49	  1) f6#(I0) -> f5#(rnd1) [y1 = 0 /\ rnd1 = rnd1]
0.00/0.49	  2) f4#(I1) -> f3#(I1) 
0.00/0.49	  3) f5#(I2) -> f4#(I2) [1 <= I2]
0.00/0.49	  4) f5#(I3) -> f1#(I3) [I3 <= 0]
0.00/0.49	  5) f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20]
0.00/0.49	  6) f3#(I5) -> f1#(I5) [20 <= I5]
0.00/0.49	
0.00/0.49	We have the following SCCs.
0.00/0.49	  { 2, 5 }
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  f4#(I1) -> f3#(I1) 
0.00/0.49	  f3#(I4) -> f4#(1 + I4) [1 + I4 <= 20]
0.00/0.49	R = 
0.00/0.49	  f7(x1) -> f6(x1) 
0.00/0.49	  f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1]
0.00/0.49	  f4(I1) -> f3(I1) 
0.00/0.49	  f5(I2) -> f4(I2) [1 <= I2]
0.00/0.49	  f5(I3) -> f1(I3) [I3 <= 0]
0.00/0.49	  f3(I4) -> f4(1 + I4) [1 + I4 <= 20]
0.00/0.49	  f3(I5) -> f1(I5) [20 <= I5]
0.00/0.49	  f1(I6) -> f2(I6) 
0.00/0.49	
0.00/0.49	We use the reverse value criterion with the projection function NU:
0.00/0.49	  NU[f3#(z1)] = 20 + -1 * (1 + z1)
0.00/0.49	  NU[f4#(z1)] = 20 + -1 * (1 + z1)
0.00/0.49	
0.00/0.49	This gives the following inequalities:
0.00/0.49	    ==>  20 + -1 * (1 + I1) >= 20 + -1 * (1 + I1)
0.00/0.49	  1 + I4 <= 20  ==>  20 + -1 * (1 + I4) > 20 + -1 * (1 + (1 + I4))  with  20 + -1 * (1 + I4) >= 0
0.00/0.49	
0.00/0.49	We remove all the strictly oriented dependency pairs.
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  f4#(I1) -> f3#(I1) 
0.00/0.49	R = 
0.00/0.49	  f7(x1) -> f6(x1) 
0.00/0.49	  f6(I0) -> f5(rnd1) [y1 = 0 /\ rnd1 = rnd1]
0.00/0.49	  f4(I1) -> f3(I1) 
0.00/0.49	  f5(I2) -> f4(I2) [1 <= I2]
0.00/0.49	  f5(I3) -> f1(I3) [I3 <= 0]
0.00/0.49	  f3(I4) -> f4(1 + I4) [1 + I4 <= 20]
0.00/0.49	  f3(I5) -> f1(I5) [20 <= I5]
0.00/0.49	  f1(I6) -> f2(I6) 
0.00/0.49	
0.00/0.49	The dependency graph for this problem is:
0.00/0.49	  2 -> 
0.00/0.49	Where:
0.00/0.49	  2) f4#(I1) -> f3#(I1) 
0.00/0.49	
0.00/0.49	We have the following SCCs.
0.00/0.49	  
0.00/3.47	EOF