1.41/1.49	MAYBE
1.41/1.49	
1.41/1.49	DP problem for innermost termination.
1.41/1.49	P =
1.41/1.49	  f6#(x1, x2) -> f5#(x1, x2) 
1.41/1.49	  f5#(I0, I1) -> f1#(I0, I1) 
1.41/1.49	  f4#(I2, I3) -> f1#(I2, I3) 
1.41/1.49	  f3#(I4, I5) -> f4#(I4, 0) 
1.41/1.49	  f2#(I6, I7) -> f3#(I6, I7) [I6 = I6]
1.41/1.49	  f1#(I8, I9) -> f2#(I8, 1) [0 <= I9]
1.41/1.49	R = 
1.41/1.49	  f6(x1, x2) -> f5(x1, x2) 
1.41/1.49	  f5(I0, I1) -> f1(I0, I1) 
1.41/1.49	  f4(I2, I3) -> f1(I2, I3) 
1.41/1.49	  f3(I4, I5) -> f4(I4, 0) 
1.41/1.49	  f2(I6, I7) -> f3(I6, I7) [I6 = I6]
1.41/1.49	  f1(I8, I9) -> f2(I8, 1) [0 <= I9]
1.41/1.49	
1.41/1.49	The dependency graph for this problem is:
1.41/1.49	  0 -> 1
1.41/1.49	  1 -> 5
1.41/1.49	  2 -> 5
1.41/1.49	  3 -> 2
1.41/1.49	  4 -> 3
1.41/1.49	  5 -> 4
1.41/1.49	Where:
1.41/1.49	  0) f6#(x1, x2) -> f5#(x1, x2) 
1.41/1.49	  1) f5#(I0, I1) -> f1#(I0, I1) 
1.41/1.49	  2) f4#(I2, I3) -> f1#(I2, I3) 
1.41/1.49	  3) f3#(I4, I5) -> f4#(I4, 0) 
1.41/1.49	  4) f2#(I6, I7) -> f3#(I6, I7) [I6 = I6]
1.41/1.49	  5) f1#(I8, I9) -> f2#(I8, 1) [0 <= I9]
1.41/1.49	
1.41/1.49	We have the following SCCs.
1.41/1.49	  { 2, 3, 4, 5 }
1.41/1.49	
1.41/1.49	DP problem for innermost termination.
1.41/1.49	P =
1.41/1.49	  f4#(I2, I3) -> f1#(I2, I3) 
1.41/1.49	  f3#(I4, I5) -> f4#(I4, 0) 
1.41/1.49	  f2#(I6, I7) -> f3#(I6, I7) [I6 = I6]
1.41/1.49	  f1#(I8, I9) -> f2#(I8, 1) [0 <= I9]
1.41/1.49	R = 
1.41/1.49	  f6(x1, x2) -> f5(x1, x2) 
1.41/1.49	  f5(I0, I1) -> f1(I0, I1) 
1.41/1.49	  f4(I2, I3) -> f1(I2, I3) 
1.41/1.49	  f3(I4, I5) -> f4(I4, 0) 
1.41/1.49	  f2(I6, I7) -> f3(I6, I7) [I6 = I6]
1.41/1.49	  f1(I8, I9) -> f2(I8, 1) [0 <= I9]
1.41/1.49	
1.41/4.47	EOF