0.64/0.68	MAYBE
0.64/0.68	
0.64/0.68	DP problem for innermost termination.
0.64/0.68	P =
0.64/0.68	  f5#(x1) -> f4#(x1) 
0.64/0.68	  f4#(I0) -> f1#(I0) 
0.64/0.68	  f3#(I1) -> f1#(I1) 
0.64/0.68	  f1#(I2) -> f3#(-1 + I2) [1 <= -1 + I2]
0.64/0.68	  f2#(I3) -> f1#(I3) 
0.64/0.68	  f1#(I4) -> f2#(-1 + I4) 
0.64/0.68	R = 
0.64/0.68	  f5(x1) -> f4(x1) 
0.64/0.68	  f4(I0) -> f1(I0) 
0.64/0.68	  f3(I1) -> f1(I1) 
0.64/0.68	  f1(I2) -> f3(-1 + I2) [1 <= -1 + I2]
0.64/0.68	  f2(I3) -> f1(I3) 
0.64/0.68	  f1(I4) -> f2(-1 + I4) 
0.64/0.68	
0.64/0.68	The dependency graph for this problem is:
0.64/0.68	  0 -> 1
0.64/0.68	  1 -> 3, 5
0.64/0.68	  2 -> 3, 5
0.64/0.68	  3 -> 2
0.64/0.68	  4 -> 3, 5
0.64/0.68	  5 -> 4
0.64/0.68	Where:
0.64/0.68	  0) f5#(x1) -> f4#(x1) 
0.64/0.68	  1) f4#(I0) -> f1#(I0) 
0.64/0.68	  2) f3#(I1) -> f1#(I1) 
0.64/0.68	  3) f1#(I2) -> f3#(-1 + I2) [1 <= -1 + I2]
0.64/0.68	  4) f2#(I3) -> f1#(I3) 
0.64/0.68	  5) f1#(I4) -> f2#(-1 + I4) 
0.64/0.68	
0.64/0.68	We have the following SCCs.
0.64/0.68	  { 2, 3, 4, 5 }
0.64/0.68	
0.64/0.68	DP problem for innermost termination.
0.64/0.68	P =
0.64/0.68	  f3#(I1) -> f1#(I1) 
0.64/0.68	  f1#(I2) -> f3#(-1 + I2) [1 <= -1 + I2]
0.64/0.68	  f2#(I3) -> f1#(I3) 
0.64/0.68	  f1#(I4) -> f2#(-1 + I4) 
0.64/0.68	R = 
0.64/0.68	  f5(x1) -> f4(x1) 
0.64/0.68	  f4(I0) -> f1(I0) 
0.64/0.68	  f3(I1) -> f1(I1) 
0.64/0.68	  f1(I2) -> f3(-1 + I2) [1 <= -1 + I2]
0.64/0.68	  f2(I3) -> f1(I3) 
0.64/0.68	  f1(I4) -> f2(-1 + I4) 
0.64/0.68	
0.64/0.68	We use the reverse value criterion with the projection function NU:
0.64/0.68	  NU[f2#(z1)] = z1
0.64/0.68	  NU[f1#(z1)] = z1
0.64/0.68	  NU[f3#(z1)] = z1
0.64/0.68	
0.64/0.68	This gives the following inequalities:
0.64/0.68	    ==>  I1 >= I1
0.64/0.68	  1 <= -1 + I2  ==>  I2 > -1 + I2  with  I2 >= 0
0.64/0.68	    ==>  I3 >= I3
0.64/0.68	    ==>  I4 >= -1 + I4
0.64/0.68	
0.64/0.68	We remove all the strictly oriented dependency pairs.
0.64/0.68	
0.64/0.68	DP problem for innermost termination.
0.64/0.68	P =
0.64/0.68	  f3#(I1) -> f1#(I1) 
0.64/0.68	  f2#(I3) -> f1#(I3) 
0.64/0.68	  f1#(I4) -> f2#(-1 + I4) 
0.64/0.68	R = 
0.64/0.68	  f5(x1) -> f4(x1) 
0.64/0.68	  f4(I0) -> f1(I0) 
0.64/0.68	  f3(I1) -> f1(I1) 
0.64/0.68	  f1(I2) -> f3(-1 + I2) [1 <= -1 + I2]
0.64/0.68	  f2(I3) -> f1(I3) 
0.64/0.68	  f1(I4) -> f2(-1 + I4) 
0.64/0.68	
0.64/0.68	The dependency graph for this problem is:
0.64/0.68	  2 -> 5
0.64/0.68	  4 -> 5
0.64/0.68	  5 -> 4
0.64/0.68	Where:
0.64/0.68	  2) f3#(I1) -> f1#(I1) 
0.64/0.68	  4) f2#(I3) -> f1#(I3) 
0.64/0.68	  5) f1#(I4) -> f2#(-1 + I4) 
0.64/0.68	
0.64/0.68	We have the following SCCs.
0.64/0.68	  { 4, 5 }
0.64/0.68	
0.64/0.68	DP problem for innermost termination.
0.64/0.68	P =
0.64/0.68	  f2#(I3) -> f1#(I3) 
0.64/0.68	  f1#(I4) -> f2#(-1 + I4) 
0.64/0.68	R = 
0.64/0.68	  f5(x1) -> f4(x1) 
0.64/0.68	  f4(I0) -> f1(I0) 
0.64/0.68	  f3(I1) -> f1(I1) 
0.64/0.68	  f1(I2) -> f3(-1 + I2) [1 <= -1 + I2]
0.64/0.68	  f2(I3) -> f1(I3) 
0.64/0.68	  f1(I4) -> f2(-1 + I4) 
0.64/0.68	
0.64/3.66	EOF