5.72/5.65	YES
5.72/5.65	
5.72/5.65	DP problem for innermost termination.
5.72/5.65	P =
5.72/5.65	  f9#(x1, x2, x3, x4) -> f8#(x1, x2, x3, x4) 
5.72/5.65	  f8#(I0, I1, I2, I3) -> f7#(0, I1, I2, I3) 
5.72/5.65	  f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
5.72/5.65	  f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) 
5.72/5.65	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) 
5.72/5.65	  f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
5.72/5.65	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I24 <= 4]
5.72/5.65	  f3#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) [4 <= I28]
5.72/5.65	  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
5.72/5.65	  f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
5.72/5.65	R = 
5.72/5.65	  f9(x1, x2, x3, x4) -> f8(x1, x2, x3, x4) 
5.72/5.65	  f8(I0, I1, I2, I3) -> f7(0, I1, I2, I3) 
5.72/5.65	  f2(I4, I5, I6, I7) -> f7(1 + I4, I5, I6, I7) 
5.72/5.65	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) 
5.72/5.65	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
5.72/5.65	  f7(I16, I17, I18, I19) -> f3(I16, I17, I18, I19) 
5.72/5.65	  f5(I20, I21, I22, I23) -> f6(I20, I21, I22, I23) 
5.72/5.65	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I24 <= 4]
5.72/5.65	  f3(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) [4 <= I28]
5.72/5.65	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) 
5.72/5.65	  f1(I36, I37, I38, I39) -> f2(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
5.72/5.65	
5.72/5.65	The dependency graph for this problem is:
5.72/5.65	  0 -> 1
5.72/5.65	  1 -> 5
5.72/5.65	  2 -> 5
5.72/5.65	  3 -> 
5.72/5.65	  4 -> 
5.72/5.65	  5 -> 6, 7
5.72/5.65	  6 -> 8, 9
5.72/5.65	  7 -> 3, 4
5.72/5.65	  8 -> 2
5.72/5.65	  9 -> 2
5.72/5.65	Where:
5.72/5.65	  0) f9#(x1, x2, x3, x4) -> f8#(x1, x2, x3, x4) 
5.72/5.65	  1) f8#(I0, I1, I2, I3) -> f7#(0, I1, I2, I3) 
5.72/5.65	  2) f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
5.72/5.65	  3) f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) 
5.72/5.65	  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) 
5.72/5.65	  5) f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
5.72/5.65	  6) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I24 <= 4]
5.72/5.65	  7) f3#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) [4 <= I28]
5.72/5.65	  8) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
5.72/5.65	  9) f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
5.72/5.65	
5.72/5.65	We have the following SCCs.
5.72/5.65	  { 2, 5, 6, 8, 9 }
5.72/5.65	
5.72/5.65	DP problem for innermost termination.
5.72/5.65	P =
5.72/5.65	  f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
5.72/5.65	  f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
5.72/5.65	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I24 <= 4]
5.72/5.65	  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
5.72/5.65	  f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
5.72/5.65	R = 
5.72/5.65	  f9(x1, x2, x3, x4) -> f8(x1, x2, x3, x4) 
5.72/5.65	  f8(I0, I1, I2, I3) -> f7(0, I1, I2, I3) 
5.72/5.65	  f2(I4, I5, I6, I7) -> f7(1 + I4, I5, I6, I7) 
5.72/5.65	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) 
5.72/5.65	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
5.72/5.65	  f7(I16, I17, I18, I19) -> f3(I16, I17, I18, I19) 
5.72/5.65	  f5(I20, I21, I22, I23) -> f6(I20, I21, I22, I23) 
5.72/5.65	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I24 <= 4]
5.72/5.65	  f3(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) [4 <= I28]
5.72/5.65	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) 
5.72/5.65	  f1(I36, I37, I38, I39) -> f2(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
5.72/5.65	
5.72/5.65	We use the extended value criterion with the projection function NU:
5.72/5.65	  NU[f1#(x0,x1,x2,x3)] = -x0 + 2
5.72/5.65	  NU[f3#(x0,x1,x2,x3)] = -x0 + 3
5.72/5.65	  NU[f7#(x0,x1,x2,x3)] = -x0 + 3
5.72/5.65	  NU[f2#(x0,x1,x2,x3)] = -x0 + 2
5.72/5.65	
5.72/5.65	This gives the following inequalities:
5.72/5.65	    ==>  -I4 + 2 >= -(1 + I4) + 3
5.72/5.65	    ==>  -I16 + 3 >= -I16 + 3
5.72/5.65	  1 + I24 <= 4  ==>  -I24 + 3 > -I24 + 2  with  -I24 + 3 >= 0
5.72/5.65	    ==>  -I32 + 2 >= -I32 + 2
5.72/5.65	  rnd2 = rnd2  ==>  -I36 + 2 >= -I36 + 2
5.72/5.65	
5.72/5.65	We remove all the strictly oriented dependency pairs.
5.72/5.65	
5.72/5.65	DP problem for innermost termination.
5.72/5.65	P =
5.72/5.65	  f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
5.72/5.65	  f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
5.72/5.65	  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
5.72/5.65	  f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
5.72/5.65	R = 
5.72/5.65	  f9(x1, x2, x3, x4) -> f8(x1, x2, x3, x4) 
5.72/5.65	  f8(I0, I1, I2, I3) -> f7(0, I1, I2, I3) 
5.72/5.65	  f2(I4, I5, I6, I7) -> f7(1 + I4, I5, I6, I7) 
5.72/5.65	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) 
5.72/5.65	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
5.72/5.65	  f7(I16, I17, I18, I19) -> f3(I16, I17, I18, I19) 
5.72/5.65	  f5(I20, I21, I22, I23) -> f6(I20, I21, I22, I23) 
5.72/5.65	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I24 <= 4]
5.72/5.65	  f3(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) [4 <= I28]
5.72/5.65	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) 
5.72/5.65	  f1(I36, I37, I38, I39) -> f2(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
5.72/5.65	
5.72/5.65	The dependency graph for this problem is:
5.72/5.65	  2 -> 5
5.72/5.65	  5 -> 
5.72/5.65	  8 -> 2
5.72/5.65	  9 -> 2
5.72/5.65	Where:
5.72/5.65	  2) f2#(I4, I5, I6, I7) -> f7#(1 + I4, I5, I6, I7) 
5.72/5.65	  5) f7#(I16, I17, I18, I19) -> f3#(I16, I17, I18, I19) 
5.72/5.65	  8) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) 
5.72/5.65	  9) f1#(I36, I37, I38, I39) -> f2#(I36, rnd2, I36, 1 + I36) [rnd2 = rnd2]
5.72/5.65	
5.72/5.65	We have the following SCCs.
5.72/5.65	  
5.72/5.66	EOF