0.82/1.38	MAYBE
0.82/1.38	
0.82/1.38	DP problem for innermost termination.
0.82/1.38	P =
0.82/1.38	  init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.82/1.38	  f2#(I0, I1, I2, I3, I4) -> f2#(I5, I1 + 1, I2 * I1, I2 * I1, I4) [I2 = I3 /\ 0 <= I5 - 1 /\ 0 <= I0 - 1 /\ I5 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ I4 <= I2 - 1]
0.82/1.38	  f2#(I6, I7, I8, I9, I10) -> f2#(I11, I7 + 1, I8 * I7, I8 * I7, I10) [I8 = I9 /\ 0 <= I11 - 1 /\ 0 <= I6 - 1 /\ I11 <= I6 /\ 0 <= I8 - 1 /\ 0 <= I7 - 1 /\ I8 <= I10 - 1]
0.82/1.38	  f1#(I12, I13, I14, I15, I16) -> f2#(I17, 1, 1, 1, I13) [0 <= I17 - 1 /\ 0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I17 <= I12]
0.82/1.38	R = 
0.82/1.38	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.82/1.38	  f2(I0, I1, I2, I3, I4) -> f2(I5, I1 + 1, I2 * I1, I2 * I1, I4) [I2 = I3 /\ 0 <= I5 - 1 /\ 0 <= I0 - 1 /\ I5 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ I4 <= I2 - 1]
0.82/1.38	  f2(I6, I7, I8, I9, I10) -> f2(I11, I7 + 1, I8 * I7, I8 * I7, I10) [I8 = I9 /\ 0 <= I11 - 1 /\ 0 <= I6 - 1 /\ I11 <= I6 /\ 0 <= I8 - 1 /\ 0 <= I7 - 1 /\ I8 <= I10 - 1]
0.82/1.38	  f1(I12, I13, I14, I15, I16) -> f2(I17, 1, 1, 1, I13) [0 <= I17 - 1 /\ 0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I17 <= I12]
0.82/1.38	
0.82/1.38	The dependency graph for this problem is:
0.82/1.38	  0 -> 3
0.82/1.38	  1 -> 1
0.82/1.38	  2 -> 1, 2
0.82/1.38	  3 -> 1, 2
0.82/1.38	Where:
0.82/1.38	  0) init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.82/1.38	  1) f2#(I0, I1, I2, I3, I4) -> f2#(I5, I1 + 1, I2 * I1, I2 * I1, I4) [I2 = I3 /\ 0 <= I5 - 1 /\ 0 <= I0 - 1 /\ I5 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ I4 <= I2 - 1]
0.82/1.38	  2) f2#(I6, I7, I8, I9, I10) -> f2#(I11, I7 + 1, I8 * I7, I8 * I7, I10) [I8 = I9 /\ 0 <= I11 - 1 /\ 0 <= I6 - 1 /\ I11 <= I6 /\ 0 <= I8 - 1 /\ 0 <= I7 - 1 /\ I8 <= I10 - 1]
0.82/1.38	  3) f1#(I12, I13, I14, I15, I16) -> f2#(I17, 1, 1, 1, I13) [0 <= I17 - 1 /\ 0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I17 <= I12]
0.82/1.38	
0.82/1.38	We have the following SCCs.
0.82/1.38	  { 2 }
0.82/1.38	  { 1 }
0.82/1.38	
0.82/1.38	DP problem for innermost termination.
0.82/1.38	P =
0.82/1.38	  f2#(I0, I1, I2, I3, I4) -> f2#(I5, I1 + 1, I2 * I1, I2 * I1, I4) [I2 = I3 /\ 0 <= I5 - 1 /\ 0 <= I0 - 1 /\ I5 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ I4 <= I2 - 1]
0.82/1.38	R = 
0.82/1.38	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.82/1.38	  f2(I0, I1, I2, I3, I4) -> f2(I5, I1 + 1, I2 * I1, I2 * I1, I4) [I2 = I3 /\ 0 <= I5 - 1 /\ 0 <= I0 - 1 /\ I5 <= I0 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ I4 <= I2 - 1]
0.82/1.38	  f2(I6, I7, I8, I9, I10) -> f2(I11, I7 + 1, I8 * I7, I8 * I7, I10) [I8 = I9 /\ 0 <= I11 - 1 /\ 0 <= I6 - 1 /\ I11 <= I6 /\ 0 <= I8 - 1 /\ 0 <= I7 - 1 /\ I8 <= I10 - 1]
0.82/1.38	  f1(I12, I13, I14, I15, I16) -> f2(I17, 1, 1, 1, I13) [0 <= I17 - 1 /\ 0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I17 <= I12]
0.82/1.38	
0.82/4.36	EOF