3.61/3.90	MAYBE
3.61/3.90	
3.61/3.90	DP problem for innermost termination.
3.61/3.90	P =
3.61/3.90	  f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 
3.61/3.90	  f9#(I0, I1, I2, I3, I4) -> f4#(0, 0, I2, rnd4, I4) [rnd4 = rnd4]
3.61/3.90	  f5#(I5, I6, I7, I8, I9) -> f4#(I5, 0, I7, I10, I9) [I9 <= 0 /\ y1 = 1 /\ I10 = I10]
3.61/3.90	  f5#(I11, I12, I13, I14, I15) -> f2#(I11, I12, I13, I14, -1 + I15) [1 <= I15]
3.61/3.90	  f8#(I16, I17, I18, I19, I20) -> f3#(I16, I17, I18, I19, I20) 
3.61/3.90	  f3#(I21, I22, I23, I24, I25) -> f8#(I21, I22, I23, I24, I25) 
3.61/3.90	  f2#(I31, I32, I33, I34, I35) -> f5#(I31, I32, I33, I34, I35) 
3.61/3.90	  f4#(I36, I37, I38, I39, I40) -> f1#(I36, I37, I38, I39, I40) 
3.61/3.90	  f1#(I41, I42, I43, I44, I45) -> f3#(I41, I42, I43, I44, I45) [1 <= I44]
3.61/3.90	  f1#(I46, I47, I48, I49, I50) -> f2#(0, I47, rnd3, I49, rnd5) [I49 <= 0 /\ I51 = 1 /\ rnd3 = rnd3 /\ rnd5 = rnd3]
3.61/3.90	R = 
3.61/3.90	  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
3.61/3.90	  f9(I0, I1, I2, I3, I4) -> f4(0, 0, I2, rnd4, I4) [rnd4 = rnd4]
3.61/3.90	  f5(I5, I6, I7, I8, I9) -> f4(I5, 0, I7, I10, I9) [I9 <= 0 /\ y1 = 1 /\ I10 = I10]
3.61/3.90	  f5(I11, I12, I13, I14, I15) -> f2(I11, I12, I13, I14, -1 + I15) [1 <= I15]
3.61/3.90	  f8(I16, I17, I18, I19, I20) -> f3(I16, I17, I18, I19, I20) 
3.61/3.90	  f3(I21, I22, I23, I24, I25) -> f8(I21, I22, I23, I24, I25) 
3.61/3.90	  f6(I26, I27, I28, I29, I30) -> f7(I26, I27, I28, I29, I30) 
3.61/3.90	  f2(I31, I32, I33, I34, I35) -> f5(I31, I32, I33, I34, I35) 
3.61/3.90	  f4(I36, I37, I38, I39, I40) -> f1(I36, I37, I38, I39, I40) 
3.61/3.90	  f1(I41, I42, I43, I44, I45) -> f3(I41, I42, I43, I44, I45) [1 <= I44]
3.61/3.90	  f1(I46, I47, I48, I49, I50) -> f2(0, I47, rnd3, I49, rnd5) [I49 <= 0 /\ I51 = 1 /\ rnd3 = rnd3 /\ rnd5 = rnd3]
3.61/3.90	
3.61/3.90	The dependency graph for this problem is:
3.61/3.90	  0 -> 1
3.61/3.90	  1 -> 7
3.61/3.90	  2 -> 7
3.61/3.90	  3 -> 6
3.61/3.90	  4 -> 5
3.61/3.90	  5 -> 4
3.61/3.90	  6 -> 2, 3
3.61/3.90	  7 -> 8, 9
3.61/3.90	  8 -> 5
3.61/3.90	  9 -> 6
3.61/3.90	Where:
3.61/3.90	  0) f10#(x1, x2, x3, x4, x5) -> f9#(x1, x2, x3, x4, x5) 
3.61/3.90	  1) f9#(I0, I1, I2, I3, I4) -> f4#(0, 0, I2, rnd4, I4) [rnd4 = rnd4]
3.61/3.90	  2) f5#(I5, I6, I7, I8, I9) -> f4#(I5, 0, I7, I10, I9) [I9 <= 0 /\ y1 = 1 /\ I10 = I10]
3.61/3.90	  3) f5#(I11, I12, I13, I14, I15) -> f2#(I11, I12, I13, I14, -1 + I15) [1 <= I15]
3.61/3.90	  4) f8#(I16, I17, I18, I19, I20) -> f3#(I16, I17, I18, I19, I20) 
3.61/3.90	  5) f3#(I21, I22, I23, I24, I25) -> f8#(I21, I22, I23, I24, I25) 
3.61/3.90	  6) f2#(I31, I32, I33, I34, I35) -> f5#(I31, I32, I33, I34, I35) 
3.61/3.90	  7) f4#(I36, I37, I38, I39, I40) -> f1#(I36, I37, I38, I39, I40) 
3.61/3.90	  8) f1#(I41, I42, I43, I44, I45) -> f3#(I41, I42, I43, I44, I45) [1 <= I44]
3.61/3.90	  9) f1#(I46, I47, I48, I49, I50) -> f2#(0, I47, rnd3, I49, rnd5) [I49 <= 0 /\ I51 = 1 /\ rnd3 = rnd3 /\ rnd5 = rnd3]
3.61/3.90	
3.61/3.90	We have the following SCCs.
3.61/3.90	  { 2, 3, 6, 7, 9 }
3.61/3.90	  { 4, 5 }
3.61/3.90	
3.61/3.90	DP problem for innermost termination.
3.61/3.90	P =
3.61/3.90	  f8#(I16, I17, I18, I19, I20) -> f3#(I16, I17, I18, I19, I20) 
3.61/3.90	  f3#(I21, I22, I23, I24, I25) -> f8#(I21, I22, I23, I24, I25) 
3.61/3.90	R = 
3.61/3.90	  f10(x1, x2, x3, x4, x5) -> f9(x1, x2, x3, x4, x5) 
3.61/3.90	  f9(I0, I1, I2, I3, I4) -> f4(0, 0, I2, rnd4, I4) [rnd4 = rnd4]
3.61/3.90	  f5(I5, I6, I7, I8, I9) -> f4(I5, 0, I7, I10, I9) [I9 <= 0 /\ y1 = 1 /\ I10 = I10]
3.61/3.90	  f5(I11, I12, I13, I14, I15) -> f2(I11, I12, I13, I14, -1 + I15) [1 <= I15]
3.61/3.90	  f8(I16, I17, I18, I19, I20) -> f3(I16, I17, I18, I19, I20) 
3.61/3.90	  f3(I21, I22, I23, I24, I25) -> f8(I21, I22, I23, I24, I25) 
3.61/3.90	  f6(I26, I27, I28, I29, I30) -> f7(I26, I27, I28, I29, I30) 
3.61/3.90	  f2(I31, I32, I33, I34, I35) -> f5(I31, I32, I33, I34, I35) 
3.61/3.90	  f4(I36, I37, I38, I39, I40) -> f1(I36, I37, I38, I39, I40) 
3.61/3.90	  f1(I41, I42, I43, I44, I45) -> f3(I41, I42, I43, I44, I45) [1 <= I44]
3.61/3.90	  f1(I46, I47, I48, I49, I50) -> f2(0, I47, rnd3, I49, rnd5) [I49 <= 0 /\ I51 = 1 /\ rnd3 = rnd3 /\ rnd5 = rnd3]
3.61/3.90	
3.61/6.88	EOF