1.42/1.47	MAYBE
1.42/1.47	
1.42/1.47	DP problem for innermost termination.
1.42/1.47	P =
1.42/1.47	  f5#(x1) -> f1#(x1) 
1.42/1.47	  f4#(I0) -> f2#(I0) 
1.42/1.47	  f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1]
1.42/1.47	  f3#(I2) -> f2#(I2) 
1.42/1.47	  f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	  f1#(I4) -> f2#(I4) 
1.42/1.47	R = 
1.42/1.47	  f5(x1) -> f1(x1) 
1.42/1.47	  f4(I0) -> f2(I0) 
1.42/1.47	  f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1]
1.42/1.47	  f3(I2) -> f2(I2) 
1.42/1.47	  f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	  f1(I4) -> f2(I4) 
1.42/1.47	
1.42/1.47	The dependency graph for this problem is:
1.42/1.47	  0 -> 5
1.42/1.47	  1 -> 2, 4
1.42/1.47	  2 -> 1
1.42/1.47	  3 -> 2, 4
1.42/1.47	  4 -> 3
1.42/1.47	  5 -> 2, 4
1.42/1.47	Where:
1.42/1.47	  0) f5#(x1) -> f1#(x1) 
1.42/1.47	  1) f4#(I0) -> f2#(I0) 
1.42/1.47	  2) f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1]
1.42/1.47	  3) f3#(I2) -> f2#(I2) 
1.42/1.47	  4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	  5) f1#(I4) -> f2#(I4) 
1.42/1.47	
1.42/1.47	We have the following SCCs.
1.42/1.47	  { 1, 2, 3, 4 }
1.42/1.47	
1.42/1.47	DP problem for innermost termination.
1.42/1.47	P =
1.42/1.47	  f4#(I0) -> f2#(I0) 
1.42/1.47	  f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1]
1.42/1.47	  f3#(I2) -> f2#(I2) 
1.42/1.47	  f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	R = 
1.42/1.47	  f5(x1) -> f1(x1) 
1.42/1.47	  f4(I0) -> f2(I0) 
1.42/1.47	  f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1]
1.42/1.47	  f3(I2) -> f2(I2) 
1.42/1.47	  f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	  f1(I4) -> f2(I4) 
1.42/1.47	
1.42/1.47	We use the reverse value criterion with the projection function NU:
1.42/1.47	  NU[f3#(z1)] = -1 + -1 + z1 + -1 * 0
1.42/1.47	  NU[f2#(z1)] = -1 + -1 + z1 + -1 * 0
1.42/1.47	  NU[f4#(z1)] = -1 + -1 + z1 + -1 * 0
1.42/1.47	
1.42/1.47	This gives the following inequalities:
1.42/1.47	    ==>  -1 + -1 + I0 + -1 * 0 >= -1 + -1 + I0 + -1 * 0
1.42/1.47	  0 <= -1 + -1 + I1  ==>  -1 + -1 + I1 + -1 * 0 > -1 + -1 + (-1 + I1) + -1 * 0  with  -1 + -1 + I1 + -1 * 0 >= 0
1.42/1.47	    ==>  -1 + -1 + I2 + -1 * 0 >= -1 + -1 + I2 + -1 * 0
1.42/1.47	  -1 + I3 <= 0  ==>  -1 + -1 + I3 + -1 * 0 >= -1 + -1 + (-1 + I3) + -1 * 0
1.42/1.47	
1.42/1.47	We remove all the strictly oriented dependency pairs.
1.42/1.47	
1.42/1.47	DP problem for innermost termination.
1.42/1.47	P =
1.42/1.47	  f4#(I0) -> f2#(I0) 
1.42/1.47	  f3#(I2) -> f2#(I2) 
1.42/1.47	  f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	R = 
1.42/1.47	  f5(x1) -> f1(x1) 
1.42/1.47	  f4(I0) -> f2(I0) 
1.42/1.47	  f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1]
1.42/1.47	  f3(I2) -> f2(I2) 
1.42/1.47	  f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	  f1(I4) -> f2(I4) 
1.42/1.47	
1.42/1.47	The dependency graph for this problem is:
1.42/1.47	  1 -> 4
1.42/1.47	  3 -> 4
1.42/1.47	  4 -> 3
1.42/1.47	Where:
1.42/1.47	  1) f4#(I0) -> f2#(I0) 
1.42/1.47	  3) f3#(I2) -> f2#(I2) 
1.42/1.47	  4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	
1.42/1.47	We have the following SCCs.
1.42/1.47	  { 3, 4 }
1.42/1.47	
1.42/1.47	DP problem for innermost termination.
1.42/1.47	P =
1.42/1.47	  f3#(I2) -> f2#(I2) 
1.42/1.47	  f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	R = 
1.42/1.47	  f5(x1) -> f1(x1) 
1.42/1.47	  f4(I0) -> f2(I0) 
1.42/1.47	  f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1]
1.42/1.47	  f3(I2) -> f2(I2) 
1.42/1.47	  f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0]
1.42/1.47	  f1(I4) -> f2(I4) 
1.42/1.47	
1.42/4.45	EOF