0.00/0.14	YES
0.00/0.14	
0.00/0.14	DP problem for innermost termination.
0.00/0.14	P =
0.00/0.14	  if#(true, I0, I1) -> pow#(I0, I1 - 1) 
0.00/0.14	  pow#(I2, I3) -> if#(I3 > 0, I2, I3) 
0.00/0.14	R = 
0.00/0.14	  if(false, x, y) -> 1 
0.00/0.14	  if(true, I0, I1) -> I0 * pow(I0, I1 - 1) 
0.00/0.14	  pow(I2, I3) -> if(I3 > 0, I2, I3) 
0.00/0.14	
0.00/0.14	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.14	
0.00/0.14	DP problem for innermost termination.
0.00/0.14	P =
0.00/0.14	  if#(true, I0, I1) -> pow#(I0, I1 - 1) 
0.00/0.14	  pow#(I2, I3) -> if#(I3 > 0, I2, I3) 
0.00/0.14	  pow#(I2, I3) -> pow#(I2, I3 - 1) [I3 > 0]
0.00/0.14	R = 
0.00/0.14	  if(false, x, y) -> 1 
0.00/0.14	  if(true, I0, I1) -> I0 * pow(I0, I1 - 1) 
0.00/0.14	  pow(I2, I3) -> if(I3 > 0, I2, I3) 
0.00/0.14	
0.00/0.14	The dependency graph for this problem is:
0.00/0.14	  0 -> 2, 1
0.00/0.14	  1 -> 
0.00/0.14	  2 -> 2, 1
0.00/0.14	Where:
0.00/0.14	  0) if#(true, I0, I1) -> pow#(I0, I1 - 1) 
0.00/0.14	  1) pow#(I2, I3) -> if#(I3 > 0, I2, I3) 
0.00/0.14	  2) pow#(I2, I3) -> pow#(I2, I3 - 1) [I3 > 0]
0.00/0.14	
0.00/0.14	We have the following SCCs.
0.00/0.14	  { 2 }
0.00/0.14	
0.00/0.14	DP problem for innermost termination.
0.00/0.14	P =
0.00/0.14	  pow#(I2, I3) -> pow#(I2, I3 - 1) [I3 > 0]
0.00/0.14	R = 
0.00/0.14	  if(false, x, y) -> 1 
0.00/0.14	  if(true, I0, I1) -> I0 * pow(I0, I1 - 1) 
0.00/0.14	  pow(I2, I3) -> if(I3 > 0, I2, I3) 
0.00/0.14	
0.00/0.14	We use the reverse value criterion with the projection function NU:
0.00/0.14	  NU[pow#(z1,z2)] = z2 + -1 * 0
0.00/0.14	
0.00/0.14	This gives the following inequalities:
0.00/0.14	  I3 > 0  ==>  I3 + -1 * 0 > I3 - 1 + -1 * 0  with  I3 + -1 * 0 >= 0
0.00/0.14	
0.00/0.14	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.12	EOF