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


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MAYBE

DP problem for innermost termination.
P =
  cu#(true, I0) -> cu#(I0 < exp(I0), I0 + 1) 
  cu#(true, I0) -> exp#(I0) 
R = 
  exp(x) -> 2 * x 
  cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) 

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

DP problem for innermost termination.
P =
  cu#(true, I0) -> cu_1#(I0) 
  cu#(true, I0) -> exp#(I0) 
  cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) 
R = 
  exp(x) -> 2 * x 
  cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) 

The dependency graph for this problem is:
  0 -> 2
  1 -> 
  2 -> 1, 0
Where:
  0) cu#(true, I0) -> cu_1#(I0) 
  1) cu#(true, I0) -> exp#(I0) 
  2) cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) 

We have the following SCCs.
  { 0, 2 }

DP problem for innermost termination.
P =
  cu#(true, I0) -> cu_1#(I0) 
  cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) 
R = 
  exp(x) -> 2 * x 
  cu(true, I0) -> cu(I0 < exp(I0), I0 + 1)