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


--------------------------------------------------------------------------------


NO

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

The system was reversed.

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

Loop of length 14 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cba.babaa
	rule cba-> aabcc at position 0
.aabcc.babaa
	rule cba-> aabcc at position 4
.aabcaabcc.baa
	rule a-> b at position 4
.aabcbabcc.baa
	rule cba-> aabcc at position 3
.aabaabccbcc.baa
	rule cba-> aabcc at position 10
.aabaabccbcaabcc.a
	rule a-> b at position 10
.aabaabccbcbabcc.a
	rule cba-> aabcc at position 9
.aabaabccbaabccbcc.a
	rule cba-> aabcc at position 7
.aabaabcaabccabccbcc.a
	rule a-> b at position 7
.aabaabcbabccabccbcc.a
	rule cc->  at position 10
.aabaabcbababccbcc.a
	rule cc->  at position 15
.aabaabcbababccb.a
	rule cba-> aabcc at position 13
.aabaabcbababcaabcc.
	rule a-> b at position 13
.aabaabcbababcbabcc.
	rule cba-> aabcc at position 12
.aabaabcbababaabccbcc.