2.40/2.43	MAYBE
2.40/2.43	
2.40/2.43	DP problem for innermost termination.
2.40/2.43	P =
2.40/2.43	  f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 
2.40/2.43	  f9#(I0, I1, I2, I3) -> f4#(0, 0, rnd3, I3) [rnd3 = rnd3]
2.40/2.43	  f5#(I4, I5, I6, I7) -> f4#(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8]
2.40/2.43	  f5#(I9, I10, I11, I12) -> f2#(I9, I10, I11, I12) [1 <= I12]
2.40/2.43	  f8#(I13, I14, I15, I16) -> f3#(I13, I14, I15, I16) 
2.40/2.43	  f3#(I17, I18, I19, I20) -> f8#(I17, I18, I19, I20) 
2.40/2.43	  f2#(I25, I26, I27, I28) -> f5#(I25, I26, I27, I28) 
2.40/2.43	  f4#(I29, I30, I31, I32) -> f1#(I29, I30, I31, I32) 
2.40/2.43	  f1#(I33, I34, I35, I36) -> f3#(I33, I34, I35, I36) [1 <= I35]
2.40/2.43	  f1#(I37, I38, I39, I40) -> f2#(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4]
2.40/2.43	R = 
2.40/2.43	  f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 
2.40/2.43	  f9(I0, I1, I2, I3) -> f4(0, 0, rnd3, I3) [rnd3 = rnd3]
2.40/2.43	  f5(I4, I5, I6, I7) -> f4(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8]
2.40/2.43	  f5(I9, I10, I11, I12) -> f2(I9, I10, I11, I12) [1 <= I12]
2.40/2.43	  f8(I13, I14, I15, I16) -> f3(I13, I14, I15, I16) 
2.40/2.43	  f3(I17, I18, I19, I20) -> f8(I17, I18, I19, I20) 
2.40/2.43	  f6(I21, I22, I23, I24) -> f7(I21, I22, I23, I24) 
2.40/2.43	  f2(I25, I26, I27, I28) -> f5(I25, I26, I27, I28) 
2.40/2.43	  f4(I29, I30, I31, I32) -> f1(I29, I30, I31, I32) 
2.40/2.43	  f1(I33, I34, I35, I36) -> f3(I33, I34, I35, I36) [1 <= I35]
2.40/2.43	  f1(I37, I38, I39, I40) -> f2(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4]
2.40/2.43	
2.40/2.43	The dependency graph for this problem is:
2.40/2.43	  0 -> 1
2.40/2.43	  1 -> 7
2.40/2.43	  2 -> 7
2.40/2.43	  3 -> 6
2.40/2.43	  4 -> 5
2.40/2.43	  5 -> 4
2.40/2.43	  6 -> 2, 3
2.40/2.43	  7 -> 8, 9
2.40/2.43	  8 -> 5
2.40/2.43	  9 -> 6
2.40/2.43	Where:
2.40/2.43	  0) f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 
2.40/2.43	  1) f9#(I0, I1, I2, I3) -> f4#(0, 0, rnd3, I3) [rnd3 = rnd3]
2.40/2.43	  2) f5#(I4, I5, I6, I7) -> f4#(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8]
2.40/2.43	  3) f5#(I9, I10, I11, I12) -> f2#(I9, I10, I11, I12) [1 <= I12]
2.40/2.43	  4) f8#(I13, I14, I15, I16) -> f3#(I13, I14, I15, I16) 
2.40/2.43	  5) f3#(I17, I18, I19, I20) -> f8#(I17, I18, I19, I20) 
2.40/2.43	  6) f2#(I25, I26, I27, I28) -> f5#(I25, I26, I27, I28) 
2.40/2.43	  7) f4#(I29, I30, I31, I32) -> f1#(I29, I30, I31, I32) 
2.40/2.43	  8) f1#(I33, I34, I35, I36) -> f3#(I33, I34, I35, I36) [1 <= I35]
2.40/2.43	  9) f1#(I37, I38, I39, I40) -> f2#(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4]
2.40/2.43	
2.40/2.43	We have the following SCCs.
2.40/2.43	  { 2, 3, 6, 7, 9 }
2.40/2.43	  { 4, 5 }
2.40/2.43	
2.40/2.43	DP problem for innermost termination.
2.40/2.43	P =
2.40/2.43	  f8#(I13, I14, I15, I16) -> f3#(I13, I14, I15, I16) 
2.40/2.43	  f3#(I17, I18, I19, I20) -> f8#(I17, I18, I19, I20) 
2.40/2.43	R = 
2.40/2.43	  f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 
2.40/2.43	  f9(I0, I1, I2, I3) -> f4(0, 0, rnd3, I3) [rnd3 = rnd3]
2.40/2.43	  f5(I4, I5, I6, I7) -> f4(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8]
2.40/2.43	  f5(I9, I10, I11, I12) -> f2(I9, I10, I11, I12) [1 <= I12]
2.40/2.43	  f8(I13, I14, I15, I16) -> f3(I13, I14, I15, I16) 
2.40/2.43	  f3(I17, I18, I19, I20) -> f8(I17, I18, I19, I20) 
2.40/2.43	  f6(I21, I22, I23, I24) -> f7(I21, I22, I23, I24) 
2.40/2.43	  f2(I25, I26, I27, I28) -> f5(I25, I26, I27, I28) 
2.40/2.43	  f4(I29, I30, I31, I32) -> f1(I29, I30, I31, I32) 
2.40/2.43	  f1(I33, I34, I35, I36) -> f3(I33, I34, I35, I36) [1 <= I35]
2.40/2.43	  f1(I37, I38, I39, I40) -> f2(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4]
2.40/2.43	
2.40/5.41	EOF