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


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NO

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

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 0 ⟶ 0 2 0 1 1 ,
  1 2 2 ⟶ 0 }

Loop of length 18 starting with a string of length 6
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.aacc
	rule ba-> acabb at position 0
.acabb.aacc
	rule ba-> acabb at position 4
.acabacabb.acc
	rule ba-> acabb at position 3
.acaacabbcabb.acc
	rule a->  at position 9
.acaacabbcbb.acc
	rule ba-> acabb at position 10
.acaacabbcbacabb.cc
	rule ba-> acabb at position 9
.acaacabbcacabbcabb.cc
	rule a->  at position 9
.acaacabbccabbcabb.cc
	rule bcc-> a at position 7
.acaacabaabbcabb.cc
	rule a->  at position 12
.acaacabaabbcbb.cc
	rule bcc-> a at position 13
.acaacabaabbcba.
	rule ba-> acabb at position 12
.acaacabaabbcacabb.
	rule a->  at position 12
.acaacabaabbccabb.
	rule bcc-> a at position 10
.acaacabaabaabb.
	rule ba-> acabb at position 9
.acaacabaaacabbabb.
	rule a->  at position 11
.acaacabaaacbbabb.
	rule ba-> acabb at position 12
.acaacabaaacbacabbbb.
	rule ba-> acabb at position 11
.acaacabaaacacabbcabbbb.
	rule a->  at position 11
.acaacabaaaccabbcabbbb.