0.00/0.33	YES
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  lif#(true, I0, I1) -> log#(I0 / I1, I1) 
0.00/0.33	  log#(I2, I3) -> lif#(I2 >= I3 && I3 > 1, I2, I3) 
0.00/0.33	R = 
0.00/0.33	  lif(false, x, y) -> 0 
0.00/0.33	  lif(true, I0, I1) -> 1 + log(I0 / I1, I1) 
0.00/0.33	  log(I2, I3) -> lif(I2 >= I3 && I3 > 1, I2, I3) 
0.00/0.33	
0.00/0.33	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  lif#(true, I0, I1) -> log#(I0 / I1, I1) 
0.00/0.33	  log#(I2, I3) -> lif#(I2 >= I3 && I3 > 1, I2, I3) 
0.00/0.33	  log#(I2, I3) -> log#(I2 / I3, I3) [I2 >= I3 && I3 > 1]
0.00/0.33	R = 
0.00/0.33	  lif(false, x, y) -> 0 
0.00/0.33	  lif(true, I0, I1) -> 1 + log(I0 / I1, I1) 
0.00/0.33	  log(I2, I3) -> lif(I2 >= I3 && I3 > 1, I2, I3) 
0.00/0.33	
0.00/0.33	The dependency graph for this problem is:
0.00/0.33	  0 -> 2, 1
0.00/0.33	  1 -> 
0.00/0.33	  2 -> 2, 1
0.00/0.33	Where:
0.00/0.33	  0) lif#(true, I0, I1) -> log#(I0 / I1, I1) 
0.00/0.33	  1) log#(I2, I3) -> lif#(I2 >= I3 && I3 > 1, I2, I3) 
0.00/0.33	  2) log#(I2, I3) -> log#(I2 / I3, I3) [I2 >= I3 && I3 > 1]
0.00/0.33	
0.00/0.33	We have the following SCCs.
0.00/0.33	  { 2 }
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  log#(I2, I3) -> log#(I2 / I3, I3) [I2 >= I3 && I3 > 1]
0.00/0.33	R = 
0.00/0.33	  lif(false, x, y) -> 0 
0.00/0.33	  lif(true, I0, I1) -> 1 + log(I0 / I1, I1) 
0.00/0.33	  log(I2, I3) -> lif(I2 >= I3 && I3 > 1, I2, I3) 
0.00/0.33	
0.00/0.33	We use the reverse value criterion with the projection function NU:
0.00/0.33	  NU[log#(z1,z2)] = z1
0.00/0.33	
0.00/0.33	This gives the following inequalities:
0.00/0.33	  I2 >= I3 && I3 > 1  ==>  I2 > I2 / I3  with  I2 >= 0
0.00/0.33	
0.00/0.33	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.31	EOF