2.41/2.87	MAYBE
2.41/2.87	
2.41/2.87	DP problem for innermost termination.
2.41/2.87	P =
2.41/2.87	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
2.41/2.87	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
2.41/2.87	  f3#(I3, I4, I5) -> f2#(I3, I6, I7) [0 = I5 /\ 0 = I4]
2.41/2.87	  f2#(I8, I9, I10) -> f3#(I8 + 1, I8 + 1, I8 + 1) [0 <= I8 - 1]
2.41/2.87	  f1#(I11, I12, I13) -> f2#(I14, I15, I16) [-1 <= y1 - 1 /\ 1 <= I12 - 1 /\ -1 <= y2 - 1 /\ 0 <= I11 - 1 /\ y1 - y2 = I14]
2.41/2.87	R = 
2.41/2.87	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
2.41/2.87	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
2.41/2.87	  f3(I3, I4, I5) -> f2(I3, I6, I7) [0 = I5 /\ 0 = I4]
2.41/2.87	  f2(I8, I9, I10) -> f3(I8 + 1, I8 + 1, I8 + 1) [0 <= I8 - 1]
2.41/2.87	  f1(I11, I12, I13) -> f2(I14, I15, I16) [-1 <= y1 - 1 /\ 1 <= I12 - 1 /\ -1 <= y2 - 1 /\ 0 <= I11 - 1 /\ y1 - y2 = I14]
2.41/2.87	
2.41/2.87	The dependency graph for this problem is:
2.41/2.87	  0 -> 4
2.41/2.87	  1 -> 1, 2
2.41/2.87	  2 -> 3
2.41/2.87	  3 -> 1
2.41/2.87	  4 -> 3
2.41/2.87	Where:
2.41/2.87	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
2.41/2.87	  1) f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
2.41/2.87	  2) f3#(I3, I4, I5) -> f2#(I3, I6, I7) [0 = I5 /\ 0 = I4]
2.41/2.87	  3) f2#(I8, I9, I10) -> f3#(I8 + 1, I8 + 1, I8 + 1) [0 <= I8 - 1]
2.41/2.87	  4) f1#(I11, I12, I13) -> f2#(I14, I15, I16) [-1 <= y1 - 1 /\ 1 <= I12 - 1 /\ -1 <= y2 - 1 /\ 0 <= I11 - 1 /\ y1 - y2 = I14]
2.41/2.87	
2.41/2.87	We have the following SCCs.
2.41/2.87	  { 1, 2, 3 }
2.41/2.87	
2.41/2.87	DP problem for innermost termination.
2.41/2.87	P =
2.41/2.87	  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
2.41/2.87	  f3#(I3, I4, I5) -> f2#(I3, I6, I7) [0 = I5 /\ 0 = I4]
2.41/2.87	  f2#(I8, I9, I10) -> f3#(I8 + 1, I8 + 1, I8 + 1) [0 <= I8 - 1]
2.41/2.87	R = 
2.41/2.87	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
2.41/2.87	  f3(I0, I1, I2) -> f3(I0, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
2.41/2.87	  f3(I3, I4, I5) -> f2(I3, I6, I7) [0 = I5 /\ 0 = I4]
2.41/2.87	  f2(I8, I9, I10) -> f3(I8 + 1, I8 + 1, I8 + 1) [0 <= I8 - 1]
2.41/2.87	  f1(I11, I12, I13) -> f2(I14, I15, I16) [-1 <= y1 - 1 /\ 1 <= I12 - 1 /\ -1 <= y2 - 1 /\ 0 <= I11 - 1 /\ y1 - y2 = I14]
2.41/2.87	
2.41/5.85	EOF