0.76/0.77	YES
0.76/0.77	
0.76/0.77	DP problem for innermost termination.
0.76/0.77	P =
0.76/0.77	  cond2#(false, x, y) -> diff#(x + 1, y) 
0.76/0.77	  cond2#(true, I0, I1) -> diff#(I0, I1 + 1) 
0.76/0.77	  cond1#(false, I2, I3) -> cond2#(I2 > I3, I2, I3) 
0.76/0.77	  diff#(I6, I7) -> cond1#(I6 = I7, I6, I7) 
0.76/0.77	R = 
0.76/0.77	  cond2(false, x, y) -> 1 + diff(x + 1, y) 
0.76/0.77	  cond2(true, I0, I1) -> 1 + diff(I0, I1 + 1) 
0.76/0.77	  cond1(false, I2, I3) -> cond2(I2 > I3, I2, I3) 
0.76/0.77	  cond1(true, I4, I5) -> 0 
0.76/0.77	  diff(I6, I7) -> cond1(I6 = I7, I6, I7) 
0.76/0.77	
0.76/0.77	This problem is converted using chaining, where edges between chained DPs are removed.
0.76/0.77	
0.76/0.77	DP problem for innermost termination.
0.76/0.77	P =
0.76/0.77	  cond2#(false, x, y) -> diff#(x + 1, y) 
0.76/0.77	  cond2#(true, I0, I1) -> diff#(I0, I1 + 1) 
0.76/0.77	  cond1#(false, I2, I3) -> cond2#(I2 > I3, I2, I3) 
0.76/0.77	  diff#(I6, I7) -> cond1#(I6 = I7, I6, I7) 
0.76/0.77	  diff#(I6, I7) -> cond2#(I6 > I7, I6, I7) [not(I6 = I7)]
0.76/0.77	  diff#(I6, I7) -> diff#(I6, I7 + 1) [not(I6 = I7), I6 > I7]
0.76/0.77	  diff#(I6, I7) -> diff#(I6 + 1, I7) [not(I6 = I7), not(I6 > I7)]
0.76/0.77	  cond1#(false, I2, I3) -> diff#(I2 + 1, I3) [not(I2 > I3)]
0.76/0.77	  cond1#(false, I2, I3) -> diff#(I2, I3 + 1) [I2 > I3]
0.76/0.77	R = 
0.76/0.77	  cond2(false, x, y) -> 1 + diff(x + 1, y) 
0.76/0.77	  cond2(true, I0, I1) -> 1 + diff(I0, I1 + 1) 
0.76/0.77	  cond1(false, I2, I3) -> cond2(I2 > I3, I2, I3) 
0.76/0.77	  cond1(true, I4, I5) -> 0 
0.76/0.77	  diff(I6, I7) -> cond1(I6 = I7, I6, I7) 
0.76/0.77	
0.76/0.77	The dependency graph for this problem is:
0.76/0.77	  0 -> 6, 5, 4, 3
0.76/0.77	  1 -> 6, 5, 4, 3
0.76/0.77	  2 -> 
0.76/0.77	  3 -> 
0.76/0.77	  4 -> 
0.76/0.77	  5 -> 5, 3, 4
0.76/0.77	  6 -> 6, 3, 4
0.76/0.77	  7 -> 3, 4, 5, 6
0.76/0.77	  8 -> 3, 4, 5
0.76/0.77	Where:
0.76/0.77	  0) cond2#(false, x, y) -> diff#(x + 1, y) 
0.76/0.77	  1) cond2#(true, I0, I1) -> diff#(I0, I1 + 1) 
0.76/0.77	  2) cond1#(false, I2, I3) -> cond2#(I2 > I3, I2, I3) 
0.76/0.77	  3) diff#(I6, I7) -> cond1#(I6 = I7, I6, I7) 
0.76/0.77	  4) diff#(I6, I7) -> cond2#(I6 > I7, I6, I7) [not(I6 = I7)]
0.76/0.77	  5) diff#(I6, I7) -> diff#(I6, I7 + 1) [not(I6 = I7), I6 > I7]
0.76/0.77	  6) diff#(I6, I7) -> diff#(I6 + 1, I7) [not(I6 = I7), not(I6 > I7)]
0.76/0.77	  7) cond1#(false, I2, I3) -> diff#(I2 + 1, I3) [not(I2 > I3)]
0.76/0.77	  8) cond1#(false, I2, I3) -> diff#(I2, I3 + 1) [I2 > I3]
0.76/0.77	
0.76/0.77	We have the following SCCs.
0.76/0.77	  { 6 }
0.76/0.77	  { 5 }
0.76/0.77	
0.76/0.77	DP problem for innermost termination.
0.76/0.77	P =
0.76/0.77	  diff#(I6, I7) -> diff#(I6, I7 + 1) [not(I6 = I7), I6 > I7]
0.76/0.77	R = 
0.76/0.77	  cond2(false, x, y) -> 1 + diff(x + 1, y) 
0.76/0.77	  cond2(true, I0, I1) -> 1 + diff(I0, I1 + 1) 
0.76/0.77	  cond1(false, I2, I3) -> cond2(I2 > I3, I2, I3) 
0.76/0.77	  cond1(true, I4, I5) -> 0 
0.76/0.77	  diff(I6, I7) -> cond1(I6 = I7, I6, I7) 
0.76/0.77	
0.76/0.77	We use the reverse value criterion with the projection function NU:
0.76/0.77	  NU[diff#(z1,z2)] = z1 + -1 * z2
0.76/0.77	
0.76/0.77	This gives the following inequalities:
0.76/0.77	  not(I6 = I7)/\I6 > I7  ==>  I6 + -1 * I7 > I6 + -1 * (I7 + 1)  with  I6 + -1 * I7 >= 0
0.76/0.77	
0.76/0.77	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.76/0.77	
0.76/0.77	DP problem for innermost termination.
0.76/0.77	P =
0.76/0.77	  diff#(I6, I7) -> diff#(I6 + 1, I7) [not(I6 = I7), not(I6 > I7)]
0.76/0.77	R = 
0.76/0.77	  cond2(false, x, y) -> 1 + diff(x + 1, y) 
0.76/0.77	  cond2(true, I0, I1) -> 1 + diff(I0, I1 + 1) 
0.76/0.77	  cond1(false, I2, I3) -> cond2(I2 > I3, I2, I3) 
0.76/0.77	  cond1(true, I4, I5) -> 0 
0.76/0.77	  diff(I6, I7) -> cond1(I6 = I7, I6, I7) 
0.76/0.77	
0.76/0.77	We use the extended value criterion with the projection function NU:
0.76/0.77	  NU[diff#(x0,x1)] = -x0 + x1
0.76/0.77	
0.76/0.77	This gives the following inequalities:
0.76/0.77	  not(I6 = I7)/\not(I6 > I7)  ==>  -I6 + I7 > -(I6 + 1) + I7  with  -I6 + I7 >= 0
0.76/0.77	
0.76/0.77	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.76/3.75	EOF