15.01/14.85	MAYBE
15.01/14.85	
15.01/14.85	DP problem for innermost termination.
15.01/14.85	P =
15.01/14.85	  f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
15.01/14.85	  f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1#(I0, rnd2, rnd3, I3, I4, I5, I6, I7, rnd9, rnd10) [rnd10 = rnd10 /\ rnd9 = rnd9 /\ y2 = y2 /\ rnd3 = rnd3 /\ y1 = y1 /\ rnd2 = rnd2]
15.01/14.85	  f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
15.01/14.85	  f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20, I21, I22, I23, I24, rnd6, rnd7, rnd8, I28, I29) [I30 = I23 /\ I31 = I24 /\ 0 <= -1 - I30 + I31 /\ rnd6 = rnd6 /\ rnd7 = rnd7 /\ y3 = I23 /\ rnd8 = rnd8]
15.01/14.85	R = 
15.01/14.85	  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
15.01/14.85	  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1(I0, rnd2, rnd3, I3, I4, I5, I6, I7, rnd9, rnd10) [rnd10 = rnd10 /\ rnd9 = rnd9 /\ y2 = y2 /\ rnd3 = rnd3 /\ y1 = y1 /\ rnd2 = rnd2]
15.01/14.85	  f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
15.01/14.85	  f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20, I21, I22, I23, I24, rnd6, rnd7, rnd8, I28, I29) [I30 = I23 /\ I31 = I24 /\ 0 <= -1 - I30 + I31 /\ rnd6 = rnd6 /\ rnd7 = rnd7 /\ y3 = I23 /\ rnd8 = rnd8]
15.01/14.85	  f1(I32, I33, I34, I35, I36, I37, I38, I39, I40, I41) -> f2(rnd1, I33, I34, I35, I36, I42, I43, I39, I40, I41) [I44 = I35 /\ I45 = I36 /\ -1 * I44 + I45 <= 0 /\ I42 = I42 /\ I43 = I43 /\ rnd1 = rnd1]
15.01/14.85	
15.01/14.85	The dependency graph for this problem is:
15.01/14.85	  0 -> 1
15.01/14.85	  1 -> 3
15.01/14.85	  2 -> 3
15.01/14.85	  3 -> 2
15.01/14.85	Where:
15.01/14.85	  0) f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
15.01/14.85	  1) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1#(I0, rnd2, rnd3, I3, I4, I5, I6, I7, rnd9, rnd10) [rnd10 = rnd10 /\ rnd9 = rnd9 /\ y2 = y2 /\ rnd3 = rnd3 /\ y1 = y1 /\ rnd2 = rnd2]
15.01/14.85	  2) f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
15.01/14.85	  3) f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20, I21, I22, I23, I24, rnd6, rnd7, rnd8, I28, I29) [I30 = I23 /\ I31 = I24 /\ 0 <= -1 - I30 + I31 /\ rnd6 = rnd6 /\ rnd7 = rnd7 /\ y3 = I23 /\ rnd8 = rnd8]
15.01/14.85	
15.01/14.85	We have the following SCCs.
15.01/14.85	  { 2, 3 }
15.01/14.85	
15.01/14.85	DP problem for innermost termination.
15.01/14.85	P =
15.01/14.85	  f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
15.01/14.85	  f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20, I21, I22, I23, I24, rnd6, rnd7, rnd8, I28, I29) [I30 = I23 /\ I31 = I24 /\ 0 <= -1 - I30 + I31 /\ rnd6 = rnd6 /\ rnd7 = rnd7 /\ y3 = I23 /\ rnd8 = rnd8]
15.01/14.85	R = 
15.01/14.85	  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
15.01/14.85	  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f1(I0, rnd2, rnd3, I3, I4, I5, I6, I7, rnd9, rnd10) [rnd10 = rnd10 /\ rnd9 = rnd9 /\ y2 = y2 /\ rnd3 = rnd3 /\ y1 = y1 /\ rnd2 = rnd2]
15.01/14.85	  f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
15.01/14.85	  f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20, I21, I22, I23, I24, rnd6, rnd7, rnd8, I28, I29) [I30 = I23 /\ I31 = I24 /\ 0 <= -1 - I30 + I31 /\ rnd6 = rnd6 /\ rnd7 = rnd7 /\ y3 = I23 /\ rnd8 = rnd8]
15.01/14.85	  f1(I32, I33, I34, I35, I36, I37, I38, I39, I40, I41) -> f2(rnd1, I33, I34, I35, I36, I42, I43, I39, I40, I41) [I44 = I35 /\ I45 = I36 /\ -1 * I44 + I45 <= 0 /\ I42 = I42 /\ I43 = I43 /\ rnd1 = rnd1]
15.01/14.85	
15.01/14.85	EOF