0.00/0.24	MAYBE
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  if#(true, I0, I1) -> exp#(I0, I1 - 1) 
0.00/0.24	  exp#(I2, I3) -> if#(I3 > 0, I2, I3) 
0.00/0.24	  cu#(true, I4) -> cu#(I4 < exp(10, 2), I4 + 1) 
0.00/0.24	  cu#(true, I4) -> exp#(10, 2) 
0.00/0.24	R = 
0.00/0.24	  if(false, x, y) -> 1 
0.00/0.24	  if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 
0.00/0.24	  exp(I2, I3) -> if(I3 > 0, I2, I3) 
0.00/0.24	  cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 
0.00/0.24	
0.00/0.24	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  if#(true, I0, I1) -> exp#(I0, I1 - 1) 
0.00/0.24	  exp#(I2, I3) -> if#(I3 > 0, I2, I3) 
0.00/0.24	  cu#(true, I4) -> cu_1#(I4) 
0.00/0.24	  cu#(true, I4) -> exp#(10, 2) 
0.00/0.24	  cu_1#(I4) -> cu#(I4 < exp(10, 2), I4 + 1) 
0.00/0.24	  exp#(I2, I3) -> exp#(I2, I3 - 1) [I3 > 0]
0.00/0.24	R = 
0.00/0.24	  if(false, x, y) -> 1 
0.00/0.24	  if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 
0.00/0.24	  exp(I2, I3) -> if(I3 > 0, I2, I3) 
0.00/0.24	  cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 
0.00/0.24	
0.00/0.24	The dependency graph for this problem is:
0.00/0.24	  0 -> 5, 1
0.00/0.24	  1 -> 
0.00/0.24	  2 -> 4
0.00/0.24	  3 -> 5, 1
0.00/0.24	  4 -> 3, 2
0.00/0.24	  5 -> 5, 1
0.00/0.24	Where:
0.00/0.24	  0) if#(true, I0, I1) -> exp#(I0, I1 - 1) 
0.00/0.24	  1) exp#(I2, I3) -> if#(I3 > 0, I2, I3) 
0.00/0.24	  2) cu#(true, I4) -> cu_1#(I4) 
0.00/0.24	  3) cu#(true, I4) -> exp#(10, 2) 
0.00/0.24	  4) cu_1#(I4) -> cu#(I4 < exp(10, 2), I4 + 1) 
0.00/0.24	  5) exp#(I2, I3) -> exp#(I2, I3 - 1) [I3 > 0]
0.00/0.24	
0.00/0.24	We have the following SCCs.
0.00/0.24	  { 2, 4 }
0.00/0.24	  { 5 }
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  exp#(I2, I3) -> exp#(I2, I3 - 1) [I3 > 0]
0.00/0.24	R = 
0.00/0.24	  if(false, x, y) -> 1 
0.00/0.24	  if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 
0.00/0.24	  exp(I2, I3) -> if(I3 > 0, I2, I3) 
0.00/0.24	  cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 
0.00/0.24	
0.00/0.24	We use the reverse value criterion with the projection function NU:
0.00/0.24	  NU[exp#(z1,z2)] = z2 + -1 * 0
0.00/0.24	
0.00/0.24	This gives the following inequalities:
0.00/0.24	  I3 > 0  ==>  I3 + -1 * 0 > I3 - 1 + -1 * 0  with  I3 + -1 * 0 >= 0
0.00/0.24	
0.00/0.24	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  cu#(true, I4) -> cu_1#(I4) 
0.00/0.24	  cu_1#(I4) -> cu#(I4 < exp(10, 2), I4 + 1) 
0.00/0.24	R = 
0.00/0.24	  if(false, x, y) -> 1 
0.00/0.24	  if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 
0.00/0.24	  exp(I2, I3) -> if(I3 > 0, I2, I3) 
0.00/0.24	  cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 
0.00/0.24	
0.00/3.22	EOF