6.34/6.30	MAYBE
6.34/6.30	
6.34/6.30	DP problem for innermost termination.
6.34/6.30	P =
6.34/6.30	  f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
6.34/6.30	  f4#(I0, I1, I2, I3) -> f2#(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	R = 
6.34/6.30	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
6.34/6.30	  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
6.34/6.30	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	The dependency graph for this problem is:
6.34/6.30	  0 -> 1
6.34/6.30	  1 -> 2
6.34/6.30	  2 -> 4, 5
6.34/6.30	  3 -> 4, 5
6.34/6.30	  4 -> 3
6.34/6.30	  5 -> 2
6.34/6.30	Where:
6.34/6.30	  0) f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
6.34/6.30	  1) f4#(I0, I1, I2, I3) -> f2#(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  2) f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  3) f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  4) f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  5) f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	We have the following SCCs.
6.34/6.30	  { 2, 3, 4, 5 }
6.34/6.30	
6.34/6.30	DP problem for innermost termination.
6.34/6.30	P =
6.34/6.30	  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	R = 
6.34/6.30	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
6.34/6.30	  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
6.34/6.30	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	We use the basic value criterion with the projection function NU:
6.34/6.30	  NU[f3#(z1,z2,z3,z4)] = z4
6.34/6.30	  NU[f1#(z1,z2,z3,z4)] = z4
6.34/6.30	  NU[f2#(z1,z2,z3,z4)] = z4
6.34/6.30	
6.34/6.30	This gives the following inequalities:
6.34/6.30	  I6 <= 0 /\ y1 = 1  ==>  I7 (>! \union =) I7
6.34/6.30	    ==>  I11 (>! \union =) I11
6.34/6.30	  1 <= I15  ==>  I15 >! -1 + I15
6.34/6.30	  I19 <= 0 /\ I21 = 0 /\ I20 = I20  ==>  I19 (>! \union =) I19
6.34/6.30	
6.34/6.30	We remove all the strictly oriented dependency pairs.
6.34/6.30	
6.34/6.30	DP problem for innermost termination.
6.34/6.30	P =
6.34/6.30	  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	R = 
6.34/6.30	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
6.34/6.30	  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
6.34/6.30	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	The dependency graph for this problem is:
6.34/6.30	  2 -> 5
6.34/6.30	  3 -> 5
6.34/6.30	  5 -> 2
6.34/6.30	Where:
6.34/6.30	  2) f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  3) f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
6.34/6.30	  5) f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/6.30	We have the following SCCs.
6.34/6.30	  { 2, 5 }
6.34/6.30	
6.34/6.30	DP problem for innermost termination.
6.34/6.30	P =
6.34/6.30	  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	R = 
6.34/6.30	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
6.34/6.30	  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
6.34/6.30	  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
6.34/6.30	  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
6.34/6.30	  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
6.34/6.30	  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
6.34/6.30	
6.34/9.28	EOF