2.95/3.01	YES
2.95/3.01	
2.95/3.01	DP problem for innermost termination.
2.95/3.01	P =
2.95/3.01	  f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
2.95/3.01	  f6#(I0, I1, I2, I3) -> f5#(I0, I1, I2, I3) 
2.95/3.01	  f6#(I4, I5, I6, I7) -> f4#(I4, I5, I6, I7) 
2.95/3.01	  f6#(I8, I9, I10, I11) -> f3#(I8, I9, I10, I11) 
2.95/3.01	  f6#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
2.95/3.01	  f6#(I20, I21, I22, I23) -> f5#(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
2.95/3.01	  f5#(I24, I25, I26, I27) -> f4#(I24, I27, I26, 0) 
2.95/3.01	  f4#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [I31 <= I28]
2.95/3.01	  f4#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [1 + I32 <= I35]
2.95/3.01	  f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 
2.95/3.01	R = 
2.95/3.01	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
2.95/3.01	  f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) 
2.95/3.01	  f6(I4, I5, I6, I7) -> f4(I4, I5, I6, I7) 
2.95/3.01	  f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) 
2.95/3.01	  f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
2.95/3.01	  f6(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 
2.95/3.01	  f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
2.95/3.01	  f5(I24, I25, I26, I27) -> f4(I24, I27, I26, 0) 
2.95/3.01	  f4(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [I31 <= I28]
2.95/3.01	  f4(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [1 + I32 <= I35]
2.95/3.01	  f3(I36, I37, I38, I39) -> f4(I36, I39, I38, 1 + I39) 
2.95/3.01	  f1(I40, I41, I42, I43) -> f2(I40, I43, I44, I45) [I45 = I44 /\ I44 = I44]
2.95/3.01	
2.95/3.01	The dependency graph for this problem is:
2.95/3.01	  0 -> 1, 2, 3, 4, 5
2.95/3.01	  1 -> 6
2.95/3.01	  2 -> 7, 8
2.95/3.01	  3 -> 9
2.95/3.01	  4 -> 
2.95/3.01	  5 -> 6
2.95/3.01	  6 -> 7, 8
2.95/3.01	  7 -> 9
2.95/3.01	  8 -> 
2.95/3.01	  9 -> 7, 8
2.95/3.01	Where:
2.95/3.01	  0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
2.95/3.01	  1) f6#(I0, I1, I2, I3) -> f5#(I0, I1, I2, I3) 
2.95/3.01	  2) f6#(I4, I5, I6, I7) -> f4#(I4, I5, I6, I7) 
2.95/3.01	  3) f6#(I8, I9, I10, I11) -> f3#(I8, I9, I10, I11) 
2.95/3.01	  4) f6#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
2.95/3.01	  5) f6#(I20, I21, I22, I23) -> f5#(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
2.95/3.01	  6) f5#(I24, I25, I26, I27) -> f4#(I24, I27, I26, 0) 
2.95/3.01	  7) f4#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [I31 <= I28]
2.95/3.01	  8) f4#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [1 + I32 <= I35]
2.95/3.01	  9) f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 
2.95/3.01	
2.95/3.01	We have the following SCCs.
2.95/3.01	  { 7, 9 }
2.95/3.01	
2.95/3.01	DP problem for innermost termination.
2.95/3.01	P =
2.95/3.01	  f4#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [I31 <= I28]
2.95/3.01	  f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 
2.95/3.01	R = 
2.95/3.01	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
2.95/3.01	  f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) 
2.95/3.01	  f6(I4, I5, I6, I7) -> f4(I4, I5, I6, I7) 
2.95/3.01	  f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) 
2.95/3.01	  f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
2.95/3.01	  f6(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 
2.95/3.01	  f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
2.95/3.01	  f5(I24, I25, I26, I27) -> f4(I24, I27, I26, 0) 
2.95/3.01	  f4(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [I31 <= I28]
2.95/3.01	  f4(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [1 + I32 <= I35]
2.95/3.01	  f3(I36, I37, I38, I39) -> f4(I36, I39, I38, 1 + I39) 
2.95/3.01	  f1(I40, I41, I42, I43) -> f2(I40, I43, I44, I45) [I45 = I44 /\ I44 = I44]
2.95/3.01	
2.95/3.01	We use the reverse value criterion with the projection function NU:
2.95/3.01	  NU[f3#(z1,z2,z3,z4)] = z1 + -1 * (1 + z4)
2.95/3.01	  NU[f4#(z1,z2,z3,z4)] = z1 + -1 * z4
2.95/3.01	
2.95/3.01	This gives the following inequalities:
2.95/3.01	  I31 <= I28  ==>  I28 + -1 * I31 > I28 + -1 * (1 + I31)  with  I28 + -1 * I31 >= 0
2.95/3.01	    ==>  I36 + -1 * (1 + I39) >= I36 + -1 * (1 + I39)
2.95/3.01	
2.95/3.01	We remove all the strictly oriented dependency pairs.
2.95/3.01	
2.95/3.01	DP problem for innermost termination.
2.95/3.01	P =
2.95/3.01	  f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 
2.95/3.01	R = 
2.95/3.01	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
2.95/3.01	  f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) 
2.95/3.01	  f6(I4, I5, I6, I7) -> f4(I4, I5, I6, I7) 
2.95/3.01	  f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) 
2.95/3.01	  f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
2.95/3.01	  f6(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 
2.95/3.01	  f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
2.95/3.01	  f5(I24, I25, I26, I27) -> f4(I24, I27, I26, 0) 
2.95/3.01	  f4(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [I31 <= I28]
2.95/3.01	  f4(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [1 + I32 <= I35]
2.95/3.01	  f3(I36, I37, I38, I39) -> f4(I36, I39, I38, 1 + I39) 
2.95/3.01	  f1(I40, I41, I42, I43) -> f2(I40, I43, I44, I45) [I45 = I44 /\ I44 = I44]
2.95/3.01	
2.95/3.01	The dependency graph for this problem is:
2.95/3.01	  9 -> 
2.95/3.01	Where:
2.95/3.01	  9) f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 
2.95/3.01	
2.95/3.01	We have the following SCCs.
2.95/3.01	  
2.95/5.99	EOF