3.91/4.22	MAYBE
3.91/4.22	
3.91/4.22	DP problem for innermost termination.
3.91/4.22	P =
3.91/4.22	  f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 
3.91/4.22	  f9#(I0, I1, I2, I3, I4) -> f1#(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
3.91/4.22	  f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [1 <= I8]
3.91/4.22	  f2#(I10, I11, I12, I13, I14) -> f3#(0, I11, I15, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ I15 = I15 /\ rnd5 = I15]
3.91/4.22	  f4#(I16, I17, I18, I19, I20) -> f1#(I16, 0, I21, I22, I20) [I20 <= 0 /\ I23 = 1 /\ I21 = I21 /\ I22 = I21]
3.91/4.22	  f4#(I24, I25, I26, I27, I28) -> f3#(I24, I25, I26, I27, I28) [1 <= I28]
3.91/4.22	  f8#(I29, I30, I31, I32, I33) -> f7#(I29, I30, I31, I32, I33) 
3.91/4.22	  f7#(I34, I35, I36, I37, I38) -> f8#(I34, I35, I36, I37, I38) 
3.91/4.22	  f3#(I44, I45, I46, I47, I48) -> f4#(I44, I45, I46, I47, I48) 
3.91/4.22	  f1#(I49, I50, I51, I52, I53) -> f2#(I49, I50, I51, I52, I53) 
3.91/4.22	R = 
3.91/4.22	  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
3.91/4.22	  f9(I0, I1, I2, I3, I4) -> f1(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
3.91/4.22	  f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [1 <= I8]
3.91/4.22	  f2(I10, I11, I12, I13, I14) -> f3(0, I11, I15, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ I15 = I15 /\ rnd5 = I15]
3.91/4.22	  f4(I16, I17, I18, I19, I20) -> f1(I16, 0, I21, I22, I20) [I20 <= 0 /\ I23 = 1 /\ I21 = I21 /\ I22 = I21]
3.91/4.22	  f4(I24, I25, I26, I27, I28) -> f3(I24, I25, I26, I27, I28) [1 <= I28]
3.91/4.22	  f8(I29, I30, I31, I32, I33) -> f7(I29, I30, I31, I32, I33) 
3.91/4.22	  f7(I34, I35, I36, I37, I38) -> f8(I34, I35, I36, I37, I38) 
3.91/4.22	  f5(I39, I40, I41, I42, I43) -> f6(I39, I40, I41, I42, I43) 
3.91/4.22	  f3(I44, I45, I46, I47, I48) -> f4(I44, I45, I46, I47, I48) 
3.91/4.22	  f1(I49, I50, I51, I52, I53) -> f2(I49, I50, I51, I52, I53) 
3.91/4.22	
3.91/4.22	The dependency graph for this problem is:
3.91/4.22	  0 -> 1
3.91/4.22	  1 -> 9
3.91/4.22	  2 -> 7
3.91/4.22	  3 -> 8
3.91/4.22	  4 -> 9
3.91/4.22	  5 -> 8
3.91/4.22	  6 -> 7
3.91/4.22	  7 -> 6
3.91/4.22	  8 -> 4, 5
3.91/4.22	  9 -> 2, 3
3.91/4.22	Where:
3.91/4.22	  0) f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 
3.91/4.22	  1) f9#(I0, I1, I2, I3, I4) -> f1#(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
3.91/4.22	  2) f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [1 <= I8]
3.91/4.22	  3) f2#(I10, I11, I12, I13, I14) -> f3#(0, I11, I15, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ I15 = I15 /\ rnd5 = I15]
3.91/4.22	  4) f4#(I16, I17, I18, I19, I20) -> f1#(I16, 0, I21, I22, I20) [I20 <= 0 /\ I23 = 1 /\ I21 = I21 /\ I22 = I21]
3.91/4.22	  5) f4#(I24, I25, I26, I27, I28) -> f3#(I24, I25, I26, I27, I28) [1 <= I28]
3.91/4.22	  6) f8#(I29, I30, I31, I32, I33) -> f7#(I29, I30, I31, I32, I33) 
3.91/4.22	  7) f7#(I34, I35, I36, I37, I38) -> f8#(I34, I35, I36, I37, I38) 
3.91/4.22	  8) f3#(I44, I45, I46, I47, I48) -> f4#(I44, I45, I46, I47, I48) 
3.91/4.22	  9) f1#(I49, I50, I51, I52, I53) -> f2#(I49, I50, I51, I52, I53) 
3.91/4.22	
3.91/4.22	We have the following SCCs.
3.91/4.22	  { 3, 4, 5, 8, 9 }
3.91/4.22	  { 6, 7 }
3.91/4.22	
3.91/4.22	DP problem for innermost termination.
3.91/4.22	P =
3.91/4.22	  f8#(I29, I30, I31, I32, I33) -> f7#(I29, I30, I31, I32, I33) 
3.91/4.22	  f7#(I34, I35, I36, I37, I38) -> f8#(I34, I35, I36, I37, I38) 
3.91/4.22	R = 
3.91/4.22	  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
3.91/4.22	  f9(I0, I1, I2, I3, I4) -> f1(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
3.91/4.22	  f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [1 <= I8]
3.91/4.22	  f2(I10, I11, I12, I13, I14) -> f3(0, I11, I15, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ I15 = I15 /\ rnd5 = I15]
3.91/4.22	  f4(I16, I17, I18, I19, I20) -> f1(I16, 0, I21, I22, I20) [I20 <= 0 /\ I23 = 1 /\ I21 = I21 /\ I22 = I21]
3.91/4.22	  f4(I24, I25, I26, I27, I28) -> f3(I24, I25, I26, I27, I28) [1 <= I28]
3.91/4.22	  f8(I29, I30, I31, I32, I33) -> f7(I29, I30, I31, I32, I33) 
3.91/4.22	  f7(I34, I35, I36, I37, I38) -> f8(I34, I35, I36, I37, I38) 
3.91/4.22	  f5(I39, I40, I41, I42, I43) -> f6(I39, I40, I41, I42, I43) 
3.91/4.22	  f3(I44, I45, I46, I47, I48) -> f4(I44, I45, I46, I47, I48) 
3.91/4.22	  f1(I49, I50, I51, I52, I53) -> f2(I49, I50, I51, I52, I53) 
3.91/4.22	
3.91/7.19	EOF