1.38/1.45	MAYBE
1.38/1.45	
1.38/1.45	DP problem for innermost termination.
1.38/1.45	P =
1.38/1.45	  f5#(x1) -> f1#(x1) 
1.38/1.45	  f4#(I0) -> f2#(I0) 
1.38/1.45	  f2#(I1) -> f4#(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	  f3#(I2) -> f2#(I2) 
1.38/1.45	  f2#(I3) -> f3#(-1 + I3) [-1 * (-1 + I3) <= 0]
1.38/1.45	  f1#(I4) -> f2#(I4) 
1.38/1.45	R = 
1.38/1.45	  f5(x1) -> f1(x1) 
1.38/1.45	  f4(I0) -> f2(I0) 
1.38/1.45	  f2(I1) -> f4(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	  f3(I2) -> f2(I2) 
1.38/1.45	  f2(I3) -> f3(-1 + I3) [-1 * (-1 + I3) <= 0]
1.38/1.45	  f1(I4) -> f2(I4) 
1.38/1.45	
1.38/1.45	The dependency graph for this problem is:
1.38/1.45	  0 -> 5
1.38/1.45	  1 -> 2, 4
1.38/1.45	  2 -> 1
1.38/1.45	  3 -> 2, 4
1.38/1.45	  4 -> 3
1.38/1.45	  5 -> 2, 4
1.38/1.45	Where:
1.38/1.45	  0) f5#(x1) -> f1#(x1) 
1.38/1.45	  1) f4#(I0) -> f2#(I0) 
1.38/1.45	  2) f2#(I1) -> f4#(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	  3) f3#(I2) -> f2#(I2) 
1.38/1.45	  4) f2#(I3) -> f3#(-1 + I3) [-1 * (-1 + I3) <= 0]
1.38/1.45	  5) f1#(I4) -> f2#(I4) 
1.38/1.45	
1.38/1.45	We have the following SCCs.
1.38/1.45	  { 1, 2, 3, 4 }
1.38/1.45	
1.38/1.45	DP problem for innermost termination.
1.38/1.45	P =
1.38/1.45	  f4#(I0) -> f2#(I0) 
1.38/1.45	  f2#(I1) -> f4#(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	  f3#(I2) -> f2#(I2) 
1.38/1.45	  f2#(I3) -> f3#(-1 + I3) [-1 * (-1 + I3) <= 0]
1.38/1.45	R = 
1.38/1.45	  f5(x1) -> f1(x1) 
1.38/1.45	  f4(I0) -> f2(I0) 
1.38/1.45	  f2(I1) -> f4(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	  f3(I2) -> f2(I2) 
1.38/1.45	  f2(I3) -> f3(-1 + I3) [-1 * (-1 + I3) <= 0]
1.38/1.45	  f1(I4) -> f2(I4) 
1.38/1.45	
1.38/1.45	We use the reverse value criterion with the projection function NU:
1.38/1.45	  NU[f3#(z1)] = z1
1.38/1.45	  NU[f2#(z1)] = z1
1.38/1.45	  NU[f4#(z1)] = z1
1.38/1.45	
1.38/1.45	This gives the following inequalities:
1.38/1.45	    ==>  I0 >= I0
1.38/1.45	  0 <= -1 - (-1 + I1)  ==>  I1 >= -1 + I1
1.38/1.45	    ==>  I2 >= I2
1.38/1.45	  -1 * (-1 + I3) <= 0  ==>  I3 > -1 + I3  with  I3 >= 0
1.38/1.45	
1.38/1.45	We remove all the strictly oriented dependency pairs.
1.38/1.45	
1.38/1.45	DP problem for innermost termination.
1.38/1.45	P =
1.38/1.45	  f4#(I0) -> f2#(I0) 
1.38/1.45	  f2#(I1) -> f4#(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	  f3#(I2) -> f2#(I2) 
1.38/1.45	R = 
1.38/1.45	  f5(x1) -> f1(x1) 
1.38/1.45	  f4(I0) -> f2(I0) 
1.38/1.45	  f2(I1) -> f4(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	  f3(I2) -> f2(I2) 
1.38/1.45	  f2(I3) -> f3(-1 + I3) [-1 * (-1 + I3) <= 0]
1.38/1.45	  f1(I4) -> f2(I4) 
1.38/1.45	
1.38/1.45	The dependency graph for this problem is:
1.38/1.45	  1 -> 2
1.38/1.45	  2 -> 1
1.38/1.45	  3 -> 2
1.38/1.45	Where:
1.38/1.45	  1) f4#(I0) -> f2#(I0) 
1.38/1.45	  2) f2#(I1) -> f4#(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	  3) f3#(I2) -> f2#(I2) 
1.38/1.45	
1.38/1.45	We have the following SCCs.
1.38/1.45	  { 1, 2 }
1.38/1.45	
1.38/1.45	DP problem for innermost termination.
1.38/1.45	P =
1.38/1.45	  f4#(I0) -> f2#(I0) 
1.38/1.45	  f2#(I1) -> f4#(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	R = 
1.38/1.45	  f5(x1) -> f1(x1) 
1.38/1.45	  f4(I0) -> f2(I0) 
1.38/1.45	  f2(I1) -> f4(-1 + I1) [0 <= -1 - (-1 + I1)]
1.38/1.45	  f3(I2) -> f2(I2) 
1.38/1.45	  f2(I3) -> f3(-1 + I3) [-1 * (-1 + I3) <= 0]
1.38/1.45	  f1(I4) -> f2(I4) 
1.38/1.45	
1.38/4.43	EOF