0.00/0.49	MAYBE
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  f6#(x1, x2) -> f5#(x1, x2) 
0.00/0.49	  f5#(I0, I1) -> f1#(I0, I1) [I0 <= I1 /\ I1 <= I0]
0.00/0.49	  f3#(I4, I5) -> f1#(I4, I5) 
0.00/0.49	  f1#(I6, I7) -> f3#(1 + I6, 1 + I7) 
0.00/0.49	R = 
0.00/0.49	  f6(x1, x2) -> f5(x1, x2) 
0.00/0.49	  f5(I0, I1) -> f1(I0, I1) [I0 <= I1 /\ I1 <= I0]
0.00/0.49	  f1(I2, I3) -> f4(I2, I3) 
0.00/0.49	  f3(I4, I5) -> f1(I4, I5) 
0.00/0.49	  f1(I6, I7) -> f3(1 + I6, 1 + I7) 
0.00/0.49	  f1(I8, I9) -> f2(I8, I9) [1 + I9 <= I8]
0.00/0.49	
0.00/0.49	The dependency graph for this problem is:
0.00/0.49	  0 -> 1
0.00/0.49	  1 -> 3
0.00/0.49	  2 -> 3
0.00/0.49	  3 -> 2
0.00/0.49	Where:
0.00/0.49	  0) f6#(x1, x2) -> f5#(x1, x2) 
0.00/0.49	  1) f5#(I0, I1) -> f1#(I0, I1) [I0 <= I1 /\ I1 <= I0]
0.00/0.49	  2) f3#(I4, I5) -> f1#(I4, I5) 
0.00/0.49	  3) f1#(I6, I7) -> f3#(1 + I6, 1 + I7) 
0.00/0.49	
0.00/0.49	We have the following SCCs.
0.00/0.49	  { 2, 3 }
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  f3#(I4, I5) -> f1#(I4, I5) 
0.00/0.49	  f1#(I6, I7) -> f3#(1 + I6, 1 + I7) 
0.00/0.49	R = 
0.00/0.49	  f6(x1, x2) -> f5(x1, x2) 
0.00/0.49	  f5(I0, I1) -> f1(I0, I1) [I0 <= I1 /\ I1 <= I0]
0.00/0.49	  f1(I2, I3) -> f4(I2, I3) 
0.00/0.49	  f3(I4, I5) -> f1(I4, I5) 
0.00/0.49	  f1(I6, I7) -> f3(1 + I6, 1 + I7) 
0.00/0.49	  f1(I8, I9) -> f2(I8, I9) [1 + I9 <= I8]
0.00/0.49	
0.00/3.47	EOF