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


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MAYBE

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

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

We have the following SCCs.
  { 2, 3, 4, 5 }

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

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1)] = z1
  NU[f1#(z1)] = z1
  NU[f3#(z1)] = z1

This gives the following inequalities:
    ==>  I1 >= I1
  1 <= -1 + I2  ==>  I2 > -1 + I2  with  I2 >= 0
    ==>  I3 >= I3
    ==>  I4 >= -1 + I4

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  2 -> 5
  4 -> 5
  5 -> 4
Where:
  2) f3#(I1) -> f1#(I1) 
  4) f2#(I3) -> f1#(I3) 
  5) f1#(I4) -> f2#(-1 + I4) 

We have the following SCCs.
  { 4, 5 }

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