3.39/3.41	MAYBE
3.39/3.41	
3.39/3.41	DP problem for innermost termination.
3.39/3.41	P =
3.39/3.41	  f6#(x1, x2) -> f1#(x1, x2) 
3.39/3.41	  f5#(I0, I1) -> f2#(I0, I1) 
3.39/3.41	  f2#(I2, I3) -> f5#(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
3.39/3.41	  f4#(I4, I5) -> f2#(I4, I5) 
3.39/3.41	  f2#(I6, I7) -> f4#(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
3.39/3.41	  f1#(I12, I13) -> f2#(I12, I13) 
3.39/3.41	R = 
3.39/3.41	  f6(x1, x2) -> f1(x1, x2) 
3.39/3.41	  f5(I0, I1) -> f2(I0, I1) 
3.39/3.41	  f2(I2, I3) -> f5(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
3.39/3.41	  f4(I4, I5) -> f2(I4, I5) 
3.39/3.41	  f2(I6, I7) -> f4(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
3.39/3.41	  f2(I10, I11) -> f3(rnd1, I11) [rnd1 = rnd1 /\ I11 <= 0 /\ -1 * I11 <= 0]
3.39/3.41	  f1(I12, I13) -> f2(I12, I13) 
3.39/3.41	
3.39/3.41	The dependency graph for this problem is:
3.39/3.41	  0 -> 5
3.39/3.41	  1 -> 2, 4
3.39/3.41	  2 -> 1
3.39/3.41	  3 -> 2, 4
3.39/3.41	  4 -> 3
3.39/3.41	  5 -> 2, 4
3.39/3.41	Where:
3.39/3.41	  0) f6#(x1, x2) -> f1#(x1, x2) 
3.39/3.41	  1) f5#(I0, I1) -> f2#(I0, I1) 
3.39/3.41	  2) f2#(I2, I3) -> f5#(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
3.39/3.41	  3) f4#(I4, I5) -> f2#(I4, I5) 
3.39/3.41	  4) f2#(I6, I7) -> f4#(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
3.39/3.41	  5) f1#(I12, I13) -> f2#(I12, I13) 
3.39/3.41	
3.39/3.41	We have the following SCCs.
3.39/3.41	  { 1, 2, 3, 4 }
3.39/3.41	
3.39/3.41	DP problem for innermost termination.
3.39/3.41	P =
3.39/3.41	  f5#(I0, I1) -> f2#(I0, I1) 
3.39/3.41	  f2#(I2, I3) -> f5#(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
3.39/3.41	  f4#(I4, I5) -> f2#(I4, I5) 
3.39/3.41	  f2#(I6, I7) -> f4#(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
3.39/3.41	R = 
3.39/3.41	  f6(x1, x2) -> f1(x1, x2) 
3.39/3.41	  f5(I0, I1) -> f2(I0, I1) 
3.39/3.41	  f2(I2, I3) -> f5(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
3.39/3.41	  f4(I4, I5) -> f2(I4, I5) 
3.39/3.41	  f2(I6, I7) -> f4(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
3.39/3.41	  f2(I10, I11) -> f3(rnd1, I11) [rnd1 = rnd1 /\ I11 <= 0 /\ -1 * I11 <= 0]
3.39/3.41	  f1(I12, I13) -> f2(I12, I13) 
3.39/3.41	
3.39/6.38	EOF