4.02/3.99	MAYBE
4.02/3.99	
4.02/3.99	DP problem for innermost termination.
4.02/3.99	P =
4.02/3.99	  f9#(x1, x2, x3, x4, x5) -> f8#(x1, x2, x3, x4, x5) 
4.02/3.99	  f8#(I0, I1, I2, I3, I4) -> f1#(I0, rnd2, rnd3, 0, 8) [rnd3 = 8 /\ 1 <= rnd2 /\ rnd2 = rnd2]
4.02/3.99	  f2#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I8, I9) [I7 <= 0]
4.02/3.99	  f2#(I10, I11, I12, I13, I14) -> f7#(rnd1, I11, I12, I13, I14) [rnd1 = rnd1 /\ 1 <= I12]
4.02/3.99	  f7#(I15, I16, I17, I18, I19) -> f1#(I15, -1 + I16, -1 + I17, 1 + I18, I19) [1 <= I15]
4.02/3.99	  f7#(I20, I21, I22, I23, I24) -> f1#(I20, I21, -1 + I22, I23, I24) [1 + I21 <= I22 /\ I20 <= 0]
4.02/3.99	  f6#(I25, I26, I27, I28, I29) -> f5#(I25, I26, I27, I28, I29) 
4.02/3.99	  f5#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 
4.02/3.99	  f1#(I40, I41, I42, I43, I44) -> f2#(I40, I41, I42, I43, I44) 
4.02/3.99	R = 
4.02/3.99	  f9(x1, x2, x3, x4, x5) -> f8(x1, x2, x3, x4, x5) 
4.02/3.99	  f8(I0, I1, I2, I3, I4) -> f1(I0, rnd2, rnd3, 0, 8) [rnd3 = 8 /\ 1 <= rnd2 /\ rnd2 = rnd2]
4.02/3.99	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, I9) [I7 <= 0]
4.02/3.99	  f2(I10, I11, I12, I13, I14) -> f7(rnd1, I11, I12, I13, I14) [rnd1 = rnd1 /\ 1 <= I12]
4.02/3.99	  f7(I15, I16, I17, I18, I19) -> f1(I15, -1 + I16, -1 + I17, 1 + I18, I19) [1 <= I15]
4.02/3.99	  f7(I20, I21, I22, I23, I24) -> f1(I20, I21, -1 + I22, I23, I24) [1 + I21 <= I22 /\ I20 <= 0]
4.02/3.99	  f6(I25, I26, I27, I28, I29) -> f5(I25, I26, I27, I28, I29) 
4.02/3.99	  f5(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) 
4.02/3.99	  f3(I35, I36, I37, I38, I39) -> f4(I35, I36, I37, I38, I39) 
4.02/3.99	  f1(I40, I41, I42, I43, I44) -> f2(I40, I41, I42, I43, I44) 
4.02/3.99	
4.02/3.99	The dependency graph for this problem is:
4.02/3.99	  0 -> 1
4.02/3.99	  1 -> 8
4.02/3.99	  2 -> 7
4.02/3.99	  3 -> 4, 5
4.02/3.99	  4 -> 8
4.02/3.99	  5 -> 8
4.02/3.99	  6 -> 7
4.02/3.99	  7 -> 6
4.02/3.99	  8 -> 2, 3
4.02/3.99	Where:
4.02/3.99	  0) f9#(x1, x2, x3, x4, x5) -> f8#(x1, x2, x3, x4, x5) 
4.02/3.99	  1) f8#(I0, I1, I2, I3, I4) -> f1#(I0, rnd2, rnd3, 0, 8) [rnd3 = 8 /\ 1 <= rnd2 /\ rnd2 = rnd2]
4.02/3.99	  2) f2#(I5, I6, I7, I8, I9) -> f5#(I5, I6, I7, I8, I9) [I7 <= 0]
4.02/3.99	  3) f2#(I10, I11, I12, I13, I14) -> f7#(rnd1, I11, I12, I13, I14) [rnd1 = rnd1 /\ 1 <= I12]
4.02/3.99	  4) f7#(I15, I16, I17, I18, I19) -> f1#(I15, -1 + I16, -1 + I17, 1 + I18, I19) [1 <= I15]
4.02/3.99	  5) f7#(I20, I21, I22, I23, I24) -> f1#(I20, I21, -1 + I22, I23, I24) [1 + I21 <= I22 /\ I20 <= 0]
4.02/3.99	  6) f6#(I25, I26, I27, I28, I29) -> f5#(I25, I26, I27, I28, I29) 
4.02/3.99	  7) f5#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 
4.02/3.99	  8) f1#(I40, I41, I42, I43, I44) -> f2#(I40, I41, I42, I43, I44) 
4.02/3.99	
4.02/3.99	We have the following SCCs.
4.02/3.99	  { 3, 4, 5, 8 }
4.02/3.99	  { 6, 7 }
4.02/3.99	
4.02/3.99	DP problem for innermost termination.
4.02/3.99	P =
4.02/3.99	  f6#(I25, I26, I27, I28, I29) -> f5#(I25, I26, I27, I28, I29) 
4.02/3.99	  f5#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 
4.02/3.99	R = 
4.02/3.99	  f9(x1, x2, x3, x4, x5) -> f8(x1, x2, x3, x4, x5) 
4.02/3.99	  f8(I0, I1, I2, I3, I4) -> f1(I0, rnd2, rnd3, 0, 8) [rnd3 = 8 /\ 1 <= rnd2 /\ rnd2 = rnd2]
4.02/3.99	  f2(I5, I6, I7, I8, I9) -> f5(I5, I6, I7, I8, I9) [I7 <= 0]
4.02/3.99	  f2(I10, I11, I12, I13, I14) -> f7(rnd1, I11, I12, I13, I14) [rnd1 = rnd1 /\ 1 <= I12]
4.02/3.99	  f7(I15, I16, I17, I18, I19) -> f1(I15, -1 + I16, -1 + I17, 1 + I18, I19) [1 <= I15]
4.02/3.99	  f7(I20, I21, I22, I23, I24) -> f1(I20, I21, -1 + I22, I23, I24) [1 + I21 <= I22 /\ I20 <= 0]
4.02/3.99	  f6(I25, I26, I27, I28, I29) -> f5(I25, I26, I27, I28, I29) 
4.02/3.99	  f5(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) 
4.02/3.99	  f3(I35, I36, I37, I38, I39) -> f4(I35, I36, I37, I38, I39) 
4.02/3.99	  f1(I40, I41, I42, I43, I44) -> f2(I40, I41, I42, I43, I44) 
4.02/3.99	
4.02/6.96	EOF