5.36/5.60	YES
5.36/5.60	
5.36/5.60	DP problem for innermost termination.
5.36/5.60	P =
5.36/5.60	  f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
5.36/5.60	  f6#(I0, I1, I2, I3) -> f3#(I0, 0, I2, rnd4) [rnd4 = rnd4]
5.36/5.60	  f4#(I4, I5, I6, I7) -> f2#(I4, I5, 1 + I6, I7) [1 + I6 <= I4]
5.36/5.60	  f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) 
5.36/5.60	  f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
5.36/5.60	  f1#(I20, I21, I22, I23) -> f3#(I20, 1 + I21, I22, I23) [1 + I21 <= I20]
5.36/5.60	  f1#(I24, I25, I26, I27) -> f2#(I24, I25, 0, I27) [I24 <= I25]
5.36/5.60	R = 
5.36/5.60	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
5.36/5.60	  f6(I0, I1, I2, I3) -> f3(I0, 0, I2, rnd4) [rnd4 = rnd4]
5.36/5.60	  f4(I4, I5, I6, I7) -> f2(I4, I5, 1 + I6, I7) [1 + I6 <= I4]
5.36/5.60	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [I8 <= I10]
5.36/5.60	  f2(I12, I13, I14, I15) -> f4(I12, I13, I14, I15) 
5.36/5.60	  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
5.36/5.60	  f1(I20, I21, I22, I23) -> f3(I20, 1 + I21, I22, I23) [1 + I21 <= I20]
5.36/5.60	  f1(I24, I25, I26, I27) -> f2(I24, I25, 0, I27) [I24 <= I25]
5.36/5.60	
5.36/5.60	The dependency graph for this problem is:
5.36/5.60	  0 -> 1
5.36/5.60	  1 -> 4
5.36/5.60	  2 -> 3
5.36/5.60	  3 -> 2
5.36/5.60	  4 -> 5, 6
5.36/5.60	  5 -> 4
5.36/5.60	  6 -> 3
5.36/5.60	Where:
5.36/5.60	  0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
5.36/5.60	  1) f6#(I0, I1, I2, I3) -> f3#(I0, 0, I2, rnd4) [rnd4 = rnd4]
5.36/5.60	  2) f4#(I4, I5, I6, I7) -> f2#(I4, I5, 1 + I6, I7) [1 + I6 <= I4]
5.36/5.60	  3) f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) 
5.36/5.60	  4) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
5.36/5.60	  5) f1#(I20, I21, I22, I23) -> f3#(I20, 1 + I21, I22, I23) [1 + I21 <= I20]
5.36/5.60	  6) f1#(I24, I25, I26, I27) -> f2#(I24, I25, 0, I27) [I24 <= I25]
5.36/5.60	
5.36/5.60	We have the following SCCs.
5.36/5.60	  { 4, 5 }
5.36/5.60	  { 2, 3 }
5.36/5.60	
5.36/5.60	DP problem for innermost termination.
5.36/5.60	P =
5.36/5.60	  f4#(I4, I5, I6, I7) -> f2#(I4, I5, 1 + I6, I7) [1 + I6 <= I4]
5.36/5.60	  f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) 
5.36/5.60	R = 
5.36/5.60	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
5.36/5.60	  f6(I0, I1, I2, I3) -> f3(I0, 0, I2, rnd4) [rnd4 = rnd4]
5.36/5.60	  f4(I4, I5, I6, I7) -> f2(I4, I5, 1 + I6, I7) [1 + I6 <= I4]
5.36/5.60	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [I8 <= I10]
5.36/5.60	  f2(I12, I13, I14, I15) -> f4(I12, I13, I14, I15) 
5.36/5.60	  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
5.36/5.60	  f1(I20, I21, I22, I23) -> f3(I20, 1 + I21, I22, I23) [1 + I21 <= I20]
5.36/5.60	  f1(I24, I25, I26, I27) -> f2(I24, I25, 0, I27) [I24 <= I25]
5.36/5.60	
5.36/5.60	We use the reverse value criterion with the projection function NU:
5.36/5.60	  NU[f2#(z1,z2,z3,z4)] = z1 + -1 * (1 + z3)
5.36/5.60	  NU[f4#(z1,z2,z3,z4)] = z1 + -1 * (1 + z3)
5.36/5.60	
5.36/5.60	This gives the following inequalities:
5.36/5.60	  1 + I6 <= I4  ==>  I4 + -1 * (1 + I6) > I4 + -1 * (1 + (1 + I6))  with  I4 + -1 * (1 + I6) >= 0
5.36/5.60	    ==>  I12 + -1 * (1 + I14) >= I12 + -1 * (1 + I14)
5.36/5.60	
5.36/5.60	We remove all the strictly oriented dependency pairs.
5.36/5.60	
5.36/5.60	DP problem for innermost termination.
5.36/5.60	P =
5.36/5.60	  f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) 
5.36/5.60	R = 
5.36/5.60	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
5.36/5.60	  f6(I0, I1, I2, I3) -> f3(I0, 0, I2, rnd4) [rnd4 = rnd4]
5.36/5.60	  f4(I4, I5, I6, I7) -> f2(I4, I5, 1 + I6, I7) [1 + I6 <= I4]
5.36/5.60	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [I8 <= I10]
5.36/5.60	  f2(I12, I13, I14, I15) -> f4(I12, I13, I14, I15) 
5.36/5.60	  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
5.36/5.60	  f1(I20, I21, I22, I23) -> f3(I20, 1 + I21, I22, I23) [1 + I21 <= I20]
5.36/5.60	  f1(I24, I25, I26, I27) -> f2(I24, I25, 0, I27) [I24 <= I25]
5.36/5.60	
5.36/5.60	The dependency graph for this problem is:
5.36/5.60	  3 -> 
5.36/5.60	Where:
5.36/5.60	  3) f2#(I12, I13, I14, I15) -> f4#(I12, I13, I14, I15) 
5.36/5.60	
5.36/5.60	We have the following SCCs.
5.36/5.60	  
5.36/5.60	
5.36/5.60	DP problem for innermost termination.
5.36/5.60	P =
5.36/5.60	  f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
5.36/5.60	  f1#(I20, I21, I22, I23) -> f3#(I20, 1 + I21, I22, I23) [1 + I21 <= I20]
5.36/5.60	R = 
5.36/5.60	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
5.36/5.60	  f6(I0, I1, I2, I3) -> f3(I0, 0, I2, rnd4) [rnd4 = rnd4]
5.36/5.60	  f4(I4, I5, I6, I7) -> f2(I4, I5, 1 + I6, I7) [1 + I6 <= I4]
5.36/5.60	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [I8 <= I10]
5.36/5.60	  f2(I12, I13, I14, I15) -> f4(I12, I13, I14, I15) 
5.36/5.60	  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
5.36/5.60	  f1(I20, I21, I22, I23) -> f3(I20, 1 + I21, I22, I23) [1 + I21 <= I20]
5.36/5.60	  f1(I24, I25, I26, I27) -> f2(I24, I25, 0, I27) [I24 <= I25]
5.36/5.60	
5.36/5.60	We use the reverse value criterion with the projection function NU:
5.36/5.60	  NU[f1#(z1,z2,z3,z4)] = z1 + -1 * (1 + z2)
5.36/5.60	  NU[f3#(z1,z2,z3,z4)] = z1 + -1 * (1 + z2)
5.36/5.60	
5.36/5.60	This gives the following inequalities:
5.36/5.60	    ==>  I16 + -1 * (1 + I17) >= I16 + -1 * (1 + I17)
5.36/5.60	  1 + I21 <= I20  ==>  I20 + -1 * (1 + I21) > I20 + -1 * (1 + (1 + I21))  with  I20 + -1 * (1 + I21) >= 0
5.36/5.60	
5.36/5.60	We remove all the strictly oriented dependency pairs.
5.36/5.60	
5.36/5.60	DP problem for innermost termination.
5.36/5.60	P =
5.36/5.60	  f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
5.36/5.60	R = 
5.36/5.60	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
5.36/5.60	  f6(I0, I1, I2, I3) -> f3(I0, 0, I2, rnd4) [rnd4 = rnd4]
5.36/5.60	  f4(I4, I5, I6, I7) -> f2(I4, I5, 1 + I6, I7) [1 + I6 <= I4]
5.36/5.60	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [I8 <= I10]
5.36/5.60	  f2(I12, I13, I14, I15) -> f4(I12, I13, I14, I15) 
5.36/5.60	  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
5.36/5.60	  f1(I20, I21, I22, I23) -> f3(I20, 1 + I21, I22, I23) [1 + I21 <= I20]
5.36/5.60	  f1(I24, I25, I26, I27) -> f2(I24, I25, 0, I27) [I24 <= I25]
5.36/5.60	
5.36/5.60	The dependency graph for this problem is:
5.36/5.60	  4 -> 
5.36/5.60	Where:
5.36/5.60	  4) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
5.36/5.60	
5.36/5.60	We have the following SCCs.
5.36/5.60	  
5.36/8.58	EOF