4.61/4.56	YES
4.61/4.56	
4.61/4.56	DP problem for innermost termination.
4.61/4.56	P =
4.61/4.56	  f5#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 
4.61/4.56	  f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
4.61/4.56	  f2#(I4, I5, I6, I7) -> f4#(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
4.61/4.56	  f1#(I12, I13, I14, I15) -> f2#(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]
4.61/4.56	R = 
4.61/4.56	  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
4.61/4.56	  f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
4.61/4.56	  f2(I4, I5, I6, I7) -> f4(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
4.61/4.56	  f2(I8, I9, I10, I11) -> f3(I8, I9, I11, I11) [10 <= I8]
4.61/4.56	  f1(I12, I13, I14, I15) -> f2(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]
4.61/4.56	
4.61/4.56	The dependency graph for this problem is:
4.61/4.56	  0 -> 3
4.61/4.56	  1 -> 2
4.61/4.56	  2 -> 1
4.61/4.56	  3 -> 2
4.61/4.56	Where:
4.61/4.56	  0) f5#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 
4.61/4.56	  1) f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
4.61/4.56	  2) f2#(I4, I5, I6, I7) -> f4#(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
4.61/4.56	  3) f1#(I12, I13, I14, I15) -> f2#(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]
4.61/4.56	
4.61/4.56	We have the following SCCs.
4.61/4.56	  { 1, 2 }
4.61/4.56	
4.61/4.56	DP problem for innermost termination.
4.61/4.56	P =
4.61/4.56	  f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
4.61/4.56	  f2#(I4, I5, I6, I7) -> f4#(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
4.61/4.56	R = 
4.61/4.56	  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
4.61/4.56	  f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
4.61/4.56	  f2(I4, I5, I6, I7) -> f4(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
4.61/4.56	  f2(I8, I9, I10, I11) -> f3(I8, I9, I11, I11) [10 <= I8]
4.61/4.56	  f1(I12, I13, I14, I15) -> f2(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]
4.61/4.56	
4.61/4.56	We use the reverse value criterion with the projection function NU:
4.61/4.56	  NU[f2#(z1,z2,z3,z4)] = 1 + z2 + -1 * (1 + z1)
4.61/4.56	  NU[f4#(z1,z2,z3,z4)] = 1 + z2 + -1 * (1 + z1)
4.61/4.56	
4.61/4.56	This gives the following inequalities:
4.61/4.56	    ==>  1 + I1 + -1 * (1 + I0) >= 1 + I1 + -1 * (1 + I0)
4.61/4.56	  1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10  ==>  1 + I5 + -1 * (1 + I4) > 1 + I5 + -1 * (1 + (1 + I4))  with  1 + I5 + -1 * (1 + I4) >= 0
4.61/4.56	
4.61/4.56	We remove all the strictly oriented dependency pairs.
4.61/4.56	
4.61/4.56	DP problem for innermost termination.
4.61/4.56	P =
4.61/4.56	  f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
4.61/4.56	R = 
4.61/4.56	  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
4.61/4.56	  f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
4.61/4.56	  f2(I4, I5, I6, I7) -> f4(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
4.61/4.56	  f2(I8, I9, I10, I11) -> f3(I8, I9, I11, I11) [10 <= I8]
4.61/4.56	  f1(I12, I13, I14, I15) -> f2(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]
4.61/4.56	
4.61/4.56	The dependency graph for this problem is:
4.61/4.56	  1 -> 
4.61/4.56	Where:
4.61/4.56	  1) f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
4.61/4.56	
4.61/4.56	We have the following SCCs.
4.61/4.56	  
4.61/7.54	EOF