3.41/3.49	MAYBE
3.41/3.49	
3.41/3.49	DP problem for innermost termination.
3.41/3.49	P =
3.41/3.49	  f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 
3.41/3.49	  f9#(I0, I1, I2, I3, I4) -> f1#(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
3.41/3.49	  f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [1 <= I8]
3.41/3.49	  f2#(I10, I11, I12, I13, I14) -> f3#(0, I11, I12, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ rnd5 = rnd5 /\ 1 <= rnd5]
3.41/3.49	  f4#(I15, I16, I17, I18, I19) -> f1#(I15, 0, I20, I21, I19) [I19 <= 0 /\ I22 = 1 /\ I20 = I20 /\ I21 = I20]
3.41/3.49	  f4#(I23, I24, I25, I26, I27) -> f3#(I23, I24, I25, I26, I27) [1 <= I27]
3.41/3.49	  f8#(I28, I29, I30, I31, I32) -> f7#(I28, I29, I30, I31, I32) 
3.41/3.49	  f7#(I33, I34, I35, I36, I37) -> f8#(I33, I34, I35, I36, I37) 
3.41/3.49	  f3#(I43, I44, I45, I46, I47) -> f4#(I43, I44, I45, I46, I47) 
3.41/3.49	  f1#(I48, I49, I50, I51, I52) -> f2#(I48, I49, I50, I51, I52) 
3.41/3.49	R = 
3.41/3.49	  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
3.41/3.49	  f9(I0, I1, I2, I3, I4) -> f1(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
3.41/3.49	  f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [1 <= I8]
3.41/3.49	  f2(I10, I11, I12, I13, I14) -> f3(0, I11, I12, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ rnd5 = rnd5 /\ 1 <= rnd5]
3.41/3.49	  f4(I15, I16, I17, I18, I19) -> f1(I15, 0, I20, I21, I19) [I19 <= 0 /\ I22 = 1 /\ I20 = I20 /\ I21 = I20]
3.41/3.49	  f4(I23, I24, I25, I26, I27) -> f3(I23, I24, I25, I26, I27) [1 <= I27]
3.41/3.49	  f8(I28, I29, I30, I31, I32) -> f7(I28, I29, I30, I31, I32) 
3.41/3.49	  f7(I33, I34, I35, I36, I37) -> f8(I33, I34, I35, I36, I37) 
3.41/3.49	  f5(I38, I39, I40, I41, I42) -> f6(I38, I39, I40, I41, I42) 
3.41/3.49	  f3(I43, I44, I45, I46, I47) -> f4(I43, I44, I45, I46, I47) 
3.41/3.49	  f1(I48, I49, I50, I51, I52) -> f2(I48, I49, I50, I51, I52) 
3.41/3.49	
3.41/3.49	The dependency graph for this problem is:
3.41/3.49	  0 -> 1
3.41/3.49	  1 -> 9
3.41/3.49	  2 -> 7
3.41/3.49	  3 -> 8
3.41/3.49	  4 -> 9
3.41/3.49	  5 -> 8
3.41/3.49	  6 -> 7
3.41/3.49	  7 -> 6
3.41/3.49	  8 -> 4, 5
3.41/3.49	  9 -> 2, 3
3.41/3.49	Where:
3.41/3.49	  0) f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 
3.41/3.49	  1) f9#(I0, I1, I2, I3, I4) -> f1#(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
3.41/3.49	  2) f2#(I5, I6, I7, I8, I9) -> f7#(I5, I6, I7, I8, I9) [1 <= I8]
3.41/3.49	  3) f2#(I10, I11, I12, I13, I14) -> f3#(0, I11, I12, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ rnd5 = rnd5 /\ 1 <= rnd5]
3.41/3.49	  4) f4#(I15, I16, I17, I18, I19) -> f1#(I15, 0, I20, I21, I19) [I19 <= 0 /\ I22 = 1 /\ I20 = I20 /\ I21 = I20]
3.41/3.49	  5) f4#(I23, I24, I25, I26, I27) -> f3#(I23, I24, I25, I26, I27) [1 <= I27]
3.41/3.49	  6) f8#(I28, I29, I30, I31, I32) -> f7#(I28, I29, I30, I31, I32) 
3.41/3.49	  7) f7#(I33, I34, I35, I36, I37) -> f8#(I33, I34, I35, I36, I37) 
3.41/3.49	  8) f3#(I43, I44, I45, I46, I47) -> f4#(I43, I44, I45, I46, I47) 
3.41/3.49	  9) f1#(I48, I49, I50, I51, I52) -> f2#(I48, I49, I50, I51, I52) 
3.41/3.49	
3.41/3.49	We have the following SCCs.
3.41/3.49	  { 3, 4, 5, 8, 9 }
3.41/3.49	  { 6, 7 }
3.41/3.49	
3.41/3.49	DP problem for innermost termination.
3.41/3.49	P =
3.41/3.49	  f8#(I28, I29, I30, I31, I32) -> f7#(I28, I29, I30, I31, I32) 
3.41/3.49	  f7#(I33, I34, I35, I36, I37) -> f8#(I33, I34, I35, I36, I37) 
3.41/3.49	R = 
3.41/3.49	  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
3.41/3.49	  f9(I0, I1, I2, I3, I4) -> f1(0, 0, rnd3, rnd4, I4) [rnd4 = rnd3 /\ rnd3 = rnd3]
3.41/3.49	  f2(I5, I6, I7, I8, I9) -> f7(I5, I6, I7, I8, I9) [1 <= I8]
3.41/3.49	  f2(I10, I11, I12, I13, I14) -> f3(0, I11, I12, I13, rnd5) [I13 <= 0 /\ y1 = 1 /\ rnd5 = rnd5 /\ 1 <= rnd5]
3.41/3.49	  f4(I15, I16, I17, I18, I19) -> f1(I15, 0, I20, I21, I19) [I19 <= 0 /\ I22 = 1 /\ I20 = I20 /\ I21 = I20]
3.41/3.49	  f4(I23, I24, I25, I26, I27) -> f3(I23, I24, I25, I26, I27) [1 <= I27]
3.41/3.49	  f8(I28, I29, I30, I31, I32) -> f7(I28, I29, I30, I31, I32) 
3.41/3.49	  f7(I33, I34, I35, I36, I37) -> f8(I33, I34, I35, I36, I37) 
3.41/3.49	  f5(I38, I39, I40, I41, I42) -> f6(I38, I39, I40, I41, I42) 
3.41/3.49	  f3(I43, I44, I45, I46, I47) -> f4(I43, I44, I45, I46, I47) 
3.41/3.49	  f1(I48, I49, I50, I51, I52) -> f2(I48, I49, I50, I51, I52) 
3.41/3.49	
3.52/6.47	EOF