1.04/1.07	MAYBE
1.04/1.07	
1.04/1.07	DP problem for innermost termination.
1.04/1.07	P =
1.04/1.07	  f5#(x1) -> f4#(x1) 
1.04/1.07	  f4#(I0) -> f1#(100) 
1.04/1.07	  f3#(I1) -> f1#(I1) 
1.04/1.07	  f2#(I2) -> f3#(I2) [301 <= I2]
1.04/1.07	  f2#(I3) -> f3#(I3) [1 + I3 <= 300]
1.04/1.07	  f1#(I4) -> f2#(-1 + I4) 
1.04/1.07	R = 
1.04/1.07	  f5(x1) -> f4(x1) 
1.04/1.07	  f4(I0) -> f1(100) 
1.04/1.07	  f3(I1) -> f1(I1) 
1.04/1.07	  f2(I2) -> f3(I2) [301 <= I2]
1.04/1.07	  f2(I3) -> f3(I3) [1 + I3 <= 300]
1.04/1.07	  f1(I4) -> f2(-1 + I4) 
1.04/1.07	
1.04/1.07	The dependency graph for this problem is:
1.04/1.07	  0 -> 1
1.04/1.07	  1 -> 5
1.04/1.07	  2 -> 5
1.04/1.07	  3 -> 2
1.04/1.07	  4 -> 2
1.04/1.07	  5 -> 3, 4
1.04/1.07	Where:
1.04/1.07	  0) f5#(x1) -> f4#(x1) 
1.04/1.07	  1) f4#(I0) -> f1#(100) 
1.04/1.07	  2) f3#(I1) -> f1#(I1) 
1.04/1.07	  3) f2#(I2) -> f3#(I2) [301 <= I2]
1.04/1.07	  4) f2#(I3) -> f3#(I3) [1 + I3 <= 300]
1.04/1.07	  5) f1#(I4) -> f2#(-1 + I4) 
1.04/1.07	
1.04/1.07	We have the following SCCs.
1.04/1.07	  { 2, 3, 4, 5 }
1.04/1.07	
1.04/1.07	DP problem for innermost termination.
1.04/1.07	P =
1.04/1.07	  f3#(I1) -> f1#(I1) 
1.04/1.07	  f2#(I2) -> f3#(I2) [301 <= I2]
1.04/1.07	  f2#(I3) -> f3#(I3) [1 + I3 <= 300]
1.04/1.07	  f1#(I4) -> f2#(-1 + I4) 
1.04/1.07	R = 
1.04/1.07	  f5(x1) -> f4(x1) 
1.04/1.07	  f4(I0) -> f1(100) 
1.04/1.07	  f3(I1) -> f1(I1) 
1.04/1.07	  f2(I2) -> f3(I2) [301 <= I2]
1.04/1.07	  f2(I3) -> f3(I3) [1 + I3 <= 300]
1.04/1.07	  f1(I4) -> f2(-1 + I4) 
1.04/1.07	
1.04/1.07	We use the extended value criterion with the projection function NU:
1.04/1.07	  NU[f2#(x0)] = x0
1.04/1.07	  NU[f1#(x0)] = x0 - 1
1.04/1.07	  NU[f3#(x0)] = x0 - 1
1.04/1.07	
1.04/1.07	This gives the following inequalities:
1.04/1.07	    ==>  I1 - 1 >= I1 - 1
1.04/1.07	  301 <= I2  ==>  I2 > I2 - 1  with  I2 >= 0
1.04/1.07	  1 + I3 <= 300  ==>  I3 >= I3 - 1
1.04/1.07	    ==>  I4 - 1 >= (-1 + I4)
1.04/1.07	
1.04/1.07	We remove all the strictly oriented dependency pairs.
1.04/1.07	
1.04/1.07	DP problem for innermost termination.
1.04/1.07	P =
1.04/1.07	  f3#(I1) -> f1#(I1) 
1.04/1.07	  f2#(I3) -> f3#(I3) [1 + I3 <= 300]
1.04/1.07	  f1#(I4) -> f2#(-1 + I4) 
1.04/1.07	R = 
1.04/1.07	  f5(x1) -> f4(x1) 
1.04/1.07	  f4(I0) -> f1(100) 
1.04/1.07	  f3(I1) -> f1(I1) 
1.04/1.07	  f2(I2) -> f3(I2) [301 <= I2]
1.04/1.07	  f2(I3) -> f3(I3) [1 + I3 <= 300]
1.04/1.07	  f1(I4) -> f2(-1 + I4) 
1.04/1.07	
1.04/1.08	EOF