SRS Standard pair #516975923

details

property	value
status	complete
benchmark	10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n026.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	MnM 3.18b
configuration	default
runtime (wallclock)	16.5827660561 seconds
cpu usage	64.530323085
max memory	8.145842176E9

stage attributes

key	value
output-size	5134
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

After renaming modulo the bijection { a ↦ 0, d ↦ 1, b ↦ 2, c ↦ 3 },
it remains to prove termination of the 7-rule system
{ 0 1 ⟶ 1 2 ,
  0 ⟶ 2 2 2 ,
  1 ⟶ ,
  0 ⟶ ,
  2 1 2 ⟶ 0 1 ,
  2 3 ⟶ 3 1 1 ,
  0 3 ⟶ 2 2 3 1 }

Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols.

After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (1,↑) ↦ 2, (2,↓) ↦ 3, (2,↑) ↦ 4, (3,↓) ↦ 5, (0,↓) ↦ 6 },
it remains to prove termination of the 19-rule system
{ 0 1 ⟶ 2 3 ,
  0 1 ⟶ 4 ,
  0 ⟶ 4 3 3 ,
  0 ⟶ 4 3 ,
  0 ⟶ 4 ,
  4 1 3 ⟶ 0 1 ,
  4 1 3 ⟶ 2 ,
  4 5 ⟶ 2 1 ,
  4 5 ⟶ 2 ,
  0 5 ⟶ 4 3 5 1 ,
  0 5 ⟶ 4 5 1 ,
  0 5 ⟶ 2 ,
  6 1 →= 1 3 ,
  6 →= 3 3 3 ,
  1 →= ,
  6 →= ,
  3 1 3 →= 6 1 ,
  3 5 →= 5 1 1 ,
  6 5 →= 3 3 5 1 }

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 2:

0 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

1 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

2 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

3 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

4 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

5 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

6 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 4 ↦ 4, 5 ↦ 5, 6 ↦ 6 },
it remains to prove termination of the 16-rule system
{ 0 1 ⟶ 2 3 ,
  0 1 ⟶ 4 ,
  0 ⟶ 4 3 3 ,
  0 ⟶ 4 3 ,
  0 ⟶ 4 ,
  4 1 3 ⟶ 0 1 ,
  4 1 3 ⟶ 2 ,
  0 5 ⟶ 4 3 5 1 ,
  0 5 ⟶ 4 5 1 ,
  6 1 →= 1 3 ,
  6 →= 3 3 3 ,
  1 →= ,
  6 →= ,
  3 1 3 →= 6 1 ,
  3 5 →= 5 1 1 ,
  6 5 →= 3 3 5 1 }

The system was filtered by the following matrix interpretation

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