3.24/3.30	YES
3.24/3.30	
3.24/3.30	DP problem for innermost termination.
3.24/3.30	P =
3.24/3.30	  f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
3.24/3.30	  f6#(I0, I1, I2) -> f3#(0, 0, rnd3) [rnd3 = rnd3 /\ y1 = 0]
3.24/3.30	  f4#(I3, I4, I5) -> f2#(I3, 1 + I4, I5) [1 + I4 <= 50]
3.24/3.30	  f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.24/3.30	  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.24/3.30	  f1#(I15, I16, I17) -> f3#(1 + I15, I16, I17) [1 + I15 <= 50]
3.24/3.30	  f1#(I18, I19, I20) -> f2#(I18, 0, I20) [50 <= I18]
3.24/3.30	R = 
3.24/3.30	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.24/3.30	  f6(I0, I1, I2) -> f3(0, 0, rnd3) [rnd3 = rnd3 /\ y1 = 0]
3.24/3.30	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 50]
3.24/3.30	  f4(I6, I7, I8) -> f5(I6, I7, I8) [50 <= I7]
3.24/3.30	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.24/3.30	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.24/3.30	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 50]
3.24/3.30	  f1(I18, I19, I20) -> f2(I18, 0, I20) [50 <= I18]
3.24/3.30	
3.24/3.30	The dependency graph for this problem is:
3.24/3.30	  0 -> 1
3.24/3.30	  1 -> 4
3.24/3.30	  2 -> 3
3.24/3.30	  3 -> 2
3.24/3.30	  4 -> 5, 6
3.24/3.30	  5 -> 4
3.24/3.30	  6 -> 3
3.24/3.30	Where:
3.24/3.30	  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
3.24/3.30	  1) f6#(I0, I1, I2) -> f3#(0, 0, rnd3) [rnd3 = rnd3 /\ y1 = 0]
3.24/3.30	  2) f4#(I3, I4, I5) -> f2#(I3, 1 + I4, I5) [1 + I4 <= 50]
3.24/3.30	  3) f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.24/3.30	  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.24/3.30	  5) f1#(I15, I16, I17) -> f3#(1 + I15, I16, I17) [1 + I15 <= 50]
3.24/3.30	  6) f1#(I18, I19, I20) -> f2#(I18, 0, I20) [50 <= I18]
3.24/3.30	
3.24/3.30	We have the following SCCs.
3.24/3.30	  { 4, 5 }
3.24/3.30	  { 2, 3 }
3.24/3.30	
3.24/3.30	DP problem for innermost termination.
3.24/3.30	P =
3.24/3.30	  f4#(I3, I4, I5) -> f2#(I3, 1 + I4, I5) [1 + I4 <= 50]
3.24/3.30	  f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.24/3.30	R = 
3.24/3.30	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.24/3.30	  f6(I0, I1, I2) -> f3(0, 0, rnd3) [rnd3 = rnd3 /\ y1 = 0]
3.24/3.30	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 50]
3.24/3.30	  f4(I6, I7, I8) -> f5(I6, I7, I8) [50 <= I7]
3.24/3.30	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.24/3.30	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.24/3.30	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 50]
3.24/3.30	  f1(I18, I19, I20) -> f2(I18, 0, I20) [50 <= I18]
3.24/3.30	
3.24/3.30	We use the reverse value criterion with the projection function NU:
3.24/3.30	  NU[f2#(z1,z2,z3)] = 50 + -1 * (1 + z2)
3.24/3.30	  NU[f4#(z1,z2,z3)] = 50 + -1 * (1 + z2)
3.24/3.30	
3.24/3.30	This gives the following inequalities:
3.24/3.30	  1 + I4 <= 50  ==>  50 + -1 * (1 + I4) > 50 + -1 * (1 + (1 + I4))  with  50 + -1 * (1 + I4) >= 0
3.24/3.30	    ==>  50 + -1 * (1 + I10) >= 50 + -1 * (1 + I10)
3.24/3.30	
3.24/3.30	We remove all the strictly oriented dependency pairs.
3.24/3.30	
3.24/3.30	DP problem for innermost termination.
3.24/3.30	P =
3.24/3.30	  f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.24/3.30	R = 
3.24/3.30	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.24/3.30	  f6(I0, I1, I2) -> f3(0, 0, rnd3) [rnd3 = rnd3 /\ y1 = 0]
3.24/3.30	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 50]
3.24/3.30	  f4(I6, I7, I8) -> f5(I6, I7, I8) [50 <= I7]
3.24/3.30	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.24/3.30	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.24/3.30	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 50]
3.24/3.30	  f1(I18, I19, I20) -> f2(I18, 0, I20) [50 <= I18]
3.24/3.30	
3.24/3.30	The dependency graph for this problem is:
3.24/3.30	  3 -> 
3.24/3.30	Where:
3.24/3.30	  3) f2#(I9, I10, I11) -> f4#(I9, I10, I11) 
3.24/3.30	
3.24/3.30	We have the following SCCs.
3.24/3.30	  
3.24/3.30	
3.24/3.30	DP problem for innermost termination.
3.24/3.30	P =
3.24/3.30	  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.24/3.30	  f1#(I15, I16, I17) -> f3#(1 + I15, I16, I17) [1 + I15 <= 50]
3.24/3.30	R = 
3.24/3.30	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.24/3.30	  f6(I0, I1, I2) -> f3(0, 0, rnd3) [rnd3 = rnd3 /\ y1 = 0]
3.24/3.30	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 50]
3.24/3.30	  f4(I6, I7, I8) -> f5(I6, I7, I8) [50 <= I7]
3.24/3.30	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.24/3.30	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.24/3.30	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 50]
3.24/3.30	  f1(I18, I19, I20) -> f2(I18, 0, I20) [50 <= I18]
3.24/3.30	
3.24/3.30	We use the reverse value criterion with the projection function NU:
3.24/3.30	  NU[f1#(z1,z2,z3)] = 50 + -1 * (1 + z1)
3.24/3.30	  NU[f3#(z1,z2,z3)] = 50 + -1 * (1 + z1)
3.24/3.30	
3.24/3.30	This gives the following inequalities:
3.24/3.30	    ==>  50 + -1 * (1 + I12) >= 50 + -1 * (1 + I12)
3.24/3.30	  1 + I15 <= 50  ==>  50 + -1 * (1 + I15) > 50 + -1 * (1 + (1 + I15))  with  50 + -1 * (1 + I15) >= 0
3.24/3.30	
3.24/3.30	We remove all the strictly oriented dependency pairs.
3.24/3.30	
3.24/3.30	DP problem for innermost termination.
3.24/3.30	P =
3.24/3.30	  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.24/3.30	R = 
3.24/3.30	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.24/3.30	  f6(I0, I1, I2) -> f3(0, 0, rnd3) [rnd3 = rnd3 /\ y1 = 0]
3.24/3.30	  f4(I3, I4, I5) -> f2(I3, 1 + I4, I5) [1 + I4 <= 50]
3.24/3.30	  f4(I6, I7, I8) -> f5(I6, I7, I8) [50 <= I7]
3.24/3.30	  f2(I9, I10, I11) -> f4(I9, I10, I11) 
3.24/3.30	  f3(I12, I13, I14) -> f1(I12, I13, I14) 
3.24/3.30	  f1(I15, I16, I17) -> f3(1 + I15, I16, I17) [1 + I15 <= 50]
3.24/3.30	  f1(I18, I19, I20) -> f2(I18, 0, I20) [50 <= I18]
3.24/3.30	
3.24/3.30	The dependency graph for this problem is:
3.24/3.30	  4 -> 
3.24/3.30	Where:
3.24/3.30	  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
3.24/3.30	
3.24/3.30	We have the following SCCs.
3.24/3.30	  
3.24/6.28	EOF