0.72/0.78	MAYBE
0.72/0.78	
0.72/0.78	DP problem for innermost termination.
0.72/0.78	P =
0.72/0.78	  f6#(x1) -> f5#(x1) 
0.72/0.78	  f5#(I0) -> f1#(I0) 
0.72/0.78	  f4#(I1) -> f1#(I1) 
0.72/0.78	  f3#(I2) -> f4#(0) 
0.72/0.78	  f2#(I3) -> f3#(I3) [1 <= I3]
0.72/0.78	  f2#(I4) -> f3#(I4) [1 + I4 <= 0]
0.72/0.78	  f1#(I5) -> f2#(1) [0 <= I5]
0.72/0.78	R = 
0.72/0.78	  f6(x1) -> f5(x1) 
0.72/0.78	  f5(I0) -> f1(I0) 
0.72/0.78	  f4(I1) -> f1(I1) 
0.72/0.78	  f3(I2) -> f4(0) 
0.72/0.78	  f2(I3) -> f3(I3) [1 <= I3]
0.72/0.78	  f2(I4) -> f3(I4) [1 + I4 <= 0]
0.72/0.78	  f1(I5) -> f2(1) [0 <= I5]
0.72/0.78	
0.72/0.78	The dependency graph for this problem is:
0.72/0.78	  0 -> 1
0.72/0.78	  1 -> 6
0.72/0.78	  2 -> 6
0.72/0.78	  3 -> 2
0.72/0.78	  4 -> 3
0.72/0.78	  5 -> 3
0.72/0.78	  6 -> 4
0.72/0.78	Where:
0.72/0.78	  0) f6#(x1) -> f5#(x1) 
0.72/0.78	  1) f5#(I0) -> f1#(I0) 
0.72/0.78	  2) f4#(I1) -> f1#(I1) 
0.72/0.78	  3) f3#(I2) -> f4#(0) 
0.72/0.78	  4) f2#(I3) -> f3#(I3) [1 <= I3]
0.72/0.78	  5) f2#(I4) -> f3#(I4) [1 + I4 <= 0]
0.72/0.78	  6) f1#(I5) -> f2#(1) [0 <= I5]
0.72/0.78	
0.72/0.78	We have the following SCCs.
0.72/0.78	  { 2, 3, 4, 6 }
0.72/0.78	
0.72/0.78	DP problem for innermost termination.
0.72/0.78	P =
0.72/0.78	  f4#(I1) -> f1#(I1) 
0.72/0.78	  f3#(I2) -> f4#(0) 
0.72/0.78	  f2#(I3) -> f3#(I3) [1 <= I3]
0.72/0.78	  f1#(I5) -> f2#(1) [0 <= I5]
0.72/0.78	R = 
0.72/0.78	  f6(x1) -> f5(x1) 
0.72/0.78	  f5(I0) -> f1(I0) 
0.72/0.78	  f4(I1) -> f1(I1) 
0.72/0.78	  f3(I2) -> f4(0) 
0.72/0.78	  f2(I3) -> f3(I3) [1 <= I3]
0.72/0.78	  f2(I4) -> f3(I4) [1 + I4 <= 0]
0.72/0.78	  f1(I5) -> f2(1) [0 <= I5]
0.72/0.78	
0.72/3.76	EOF