/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files


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YES

DP problem for innermost termination.
P =
  cond2#(false, x, y) -> diff#(x + 1, y) 
  cond2#(true, I0, I1) -> diff#(I0, I1 + 1) 
  cond1#(false, I2, I3) -> cond2#(I2 > I3, I2, I3) 
  diff#(I6, I7) -> cond1#(I6 = I7, I6, I7) 
R = 
  cond2(false, x, y) -> 1 + diff(x + 1, y) 
  cond2(true, I0, I1) -> 1 + diff(I0, I1 + 1) 
  cond1(false, I2, I3) -> cond2(I2 > I3, I2, I3) 
  cond1(true, I4, I5) -> 0 
  diff(I6, I7) -> cond1(I6 = I7, I6, I7) 

This problem is converted using chaining, where edges between chained DPs are removed.

DP problem for innermost termination.
P =
  cond2#(false, x, y) -> diff#(x + 1, y) 
  cond2#(true, I0, I1) -> diff#(I0, I1 + 1) 
  cond1#(false, I2, I3) -> cond2#(I2 > I3, I2, I3) 
  diff#(I6, I7) -> cond1#(I6 = I7, I6, I7) 
  diff#(I6, I7) -> cond2#(I6 > I7, I6, I7) [not(I6 = I7)]
  diff#(I6, I7) -> diff#(I6, I7 + 1) [not(I6 = I7), I6 > I7]
  diff#(I6, I7) -> diff#(I6 + 1, I7) [not(I6 = I7), not(I6 > I7)]
  cond1#(false, I2, I3) -> diff#(I2 + 1, I3) [not(I2 > I3)]
  cond1#(false, I2, I3) -> diff#(I2, I3 + 1) [I2 > I3]
R = 
  cond2(false, x, y) -> 1 + diff(x + 1, y) 
  cond2(true, I0, I1) -> 1 + diff(I0, I1 + 1) 
  cond1(false, I2, I3) -> cond2(I2 > I3, I2, I3) 
  cond1(true, I4, I5) -> 0 
  diff(I6, I7) -> cond1(I6 = I7, I6, I7) 

The dependency graph for this problem is:
  0 -> 6, 5, 4, 3
  1 -> 6, 5, 4, 3
  2 -> 
  3 -> 
  4 -> 
  5 -> 5, 3, 4
  6 -> 6, 3, 4
  7 -> 3, 4, 5, 6
  8 -> 3, 4, 5
Where:
  0) cond2#(false, x, y) -> diff#(x + 1, y) 
  1) cond2#(true, I0, I1) -> diff#(I0, I1 + 1) 
  2) cond1#(false, I2, I3) -> cond2#(I2 > I3, I2, I3) 
  3) diff#(I6, I7) -> cond1#(I6 = I7, I6, I7) 
  4) diff#(I6, I7) -> cond2#(I6 > I7, I6, I7) [not(I6 = I7)]
  5) diff#(I6, I7) -> diff#(I6, I7 + 1) [not(I6 = I7), I6 > I7]
  6) diff#(I6, I7) -> diff#(I6 + 1, I7) [not(I6 = I7), not(I6 > I7)]
  7) cond1#(false, I2, I3) -> diff#(I2 + 1, I3) [not(I2 > I3)]
  8) cond1#(false, I2, I3) -> diff#(I2, I3 + 1) [I2 > I3]

We have the following SCCs.
  { 6 }
  { 5 }

DP problem for innermost termination.
P =
  diff#(I6, I7) -> diff#(I6, I7 + 1) [not(I6 = I7), I6 > I7]
R = 
  cond2(false, x, y) -> 1 + diff(x + 1, y) 
  cond2(true, I0, I1) -> 1 + diff(I0, I1 + 1) 
  cond1(false, I2, I3) -> cond2(I2 > I3, I2, I3) 
  cond1(true, I4, I5) -> 0 
  diff(I6, I7) -> cond1(I6 = I7, I6, I7) 

We use the reverse value criterion with the projection function NU:
  NU[diff#(z1,z2)] = z1 + -1 * z2

This gives the following inequalities:
  not(I6 = I7)/\I6 > I7  ==>  I6 + -1 * I7 > I6 + -1 * (I7 + 1)  with  I6 + -1 * I7 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.

DP problem for innermost termination.
P =
  diff#(I6, I7) -> diff#(I6 + 1, I7) [not(I6 = I7), not(I6 > I7)]
R = 
  cond2(false, x, y) -> 1 + diff(x + 1, y) 
  cond2(true, I0, I1) -> 1 + diff(I0, I1 + 1) 
  cond1(false, I2, I3) -> cond2(I2 > I3, I2, I3) 
  cond1(true, I4, I5) -> 0 
  diff(I6, I7) -> cond1(I6 = I7, I6, I7) 

We use the extended value criterion with the projection function NU:
  NU[diff#(x0,x1)] = -x0 + x1

This gives the following inequalities:
  not(I6 = I7)/\not(I6 > I7)  ==>  -I6 + I7 > -(I6 + 1) + I7  with  -I6 + I7 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.