/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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MAYBE

DP problem for innermost termination.
P =
  f5#(x1) -> f1#(x1) 
  f4#(I0) -> f2#(I0) 
  f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1]
  f3#(I2) -> f2#(I2) 
  f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
  f1#(I4) -> f2#(I4) 
R = 
  f5(x1) -> f1(x1) 
  f4(I0) -> f2(I0) 
  f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1]
  f3(I2) -> f2(I2) 
  f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0]
  f1(I4) -> f2(I4) 

The dependency graph for this problem is:
  0 -> 5
  1 -> 2, 4
  2 -> 1
  3 -> 2, 4
  4 -> 3
  5 -> 2, 4
Where:
  0) f5#(x1) -> f1#(x1) 
  1) f4#(I0) -> f2#(I0) 
  2) f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1]
  3) f3#(I2) -> f2#(I2) 
  4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
  5) f1#(I4) -> f2#(I4) 

We have the following SCCs.
  { 1, 2, 3, 4 }

DP problem for innermost termination.
P =
  f4#(I0) -> f2#(I0) 
  f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1]
  f3#(I2) -> f2#(I2) 
  f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
R = 
  f5(x1) -> f1(x1) 
  f4(I0) -> f2(I0) 
  f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1]
  f3(I2) -> f2(I2) 
  f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0]
  f1(I4) -> f2(I4) 

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1)] = -1 + -1 + z1 + -1 * 0
  NU[f2#(z1)] = -1 + -1 + z1 + -1 * 0
  NU[f4#(z1)] = -1 + -1 + z1 + -1 * 0

This gives the following inequalities:
    ==>  -1 + -1 + I0 + -1 * 0 >= -1 + -1 + I0 + -1 * 0
  0 <= -1 + -1 + I1  ==>  -1 + -1 + I1 + -1 * 0 > -1 + -1 + (-1 + I1) + -1 * 0  with  -1 + -1 + I1 + -1 * 0 >= 0
    ==>  -1 + -1 + I2 + -1 * 0 >= -1 + -1 + I2 + -1 * 0
  -1 + I3 <= 0  ==>  -1 + -1 + I3 + -1 * 0 >= -1 + -1 + (-1 + I3) + -1 * 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I0) -> f2#(I0) 
  f3#(I2) -> f2#(I2) 
  f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
R = 
  f5(x1) -> f1(x1) 
  f4(I0) -> f2(I0) 
  f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1]
  f3(I2) -> f2(I2) 
  f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0]
  f1(I4) -> f2(I4) 

The dependency graph for this problem is:
  1 -> 4
  3 -> 4
  4 -> 3
Where:
  1) f4#(I0) -> f2#(I0) 
  3) f3#(I2) -> f2#(I2) 
  4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]

We have the following SCCs.
  { 3, 4 }

DP problem for innermost termination.
P =
  f3#(I2) -> f2#(I2) 
  f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0]
R = 
  f5(x1) -> f1(x1) 
  f4(I0) -> f2(I0) 
  f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1]
  f3(I2) -> f2(I2) 
  f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0]
  f1(I4) -> f2(I4)