0.00/0.12	YES
0.00/0.12	
0.00/0.12	DP problem for innermost termination.
0.00/0.12	P =
0.00/0.12	  mult#(x, y) -> mult#(0 - x, y) [0 > x]
0.00/0.12	  mult#(I0, I1) -> mult#(I0 - 1, I1) [I0 > 0]
0.00/0.12	R = 
0.00/0.12	  mult(x, y) -> 0 - mult(0 - x, y) [0 > x]
0.00/0.12	  mult(I0, I1) -> mult(I0 - 1, I1) + I1 [I0 > 0]
0.00/0.12	  mult(0, I2) -> 0 
0.00/0.12	
0.00/0.12	The dependency graph for this problem is:
0.00/0.12	  0 -> 1
0.00/0.12	  1 -> 1
0.00/0.12	Where:
0.00/0.12	  0) mult#(x, y) -> mult#(0 - x, y) [0 > x]
0.00/0.12	  1) mult#(I0, I1) -> mult#(I0 - 1, I1) [I0 > 0]
0.00/0.12	
0.00/0.12	We have the following SCCs.
0.00/0.12	  { 1 }
0.00/0.12	
0.00/0.12	DP problem for innermost termination.
0.00/0.12	P =
0.00/0.12	  mult#(I0, I1) -> mult#(I0 - 1, I1) [I0 > 0]
0.00/0.12	R = 
0.00/0.12	  mult(x, y) -> 0 - mult(0 - x, y) [0 > x]
0.00/0.12	  mult(I0, I1) -> mult(I0 - 1, I1) + I1 [I0 > 0]
0.00/0.12	  mult(0, I2) -> 0 
0.00/0.12	
0.00/0.12	We use the reverse value criterion with the projection function NU:
0.00/0.12	  NU[mult#(z1,z2)] = z1
0.00/0.12	
0.00/0.12	This gives the following inequalities:
0.00/0.12	  I0 > 0  ==>  I0 > I0 - 1  with  I0 >= 0
0.00/0.12	
0.00/0.12	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.10	EOF