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


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NO

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

Loop of length 19 starting with a string of length 7
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.aaa.cccc
	rule aaa-> bbc at position 0
.bbc.cccc
	rule ccc-> bbb at position 2
.bbbbb.cc
	rule bbb-> acc at position 2
.bbacc.cc
	rule ccc-> bbb at position 3
.bbabbb.c
	rule bbb-> acc at position 3
.bbaacc.c
	rule ccc-> bbb at position 4
.bbaabbb.
	rule bbb-> acc at position 4
.bbaaacc.
	rule aaa-> bbc at position 2
.bbbbccc.
	rule bbb-> acc at position 1
.baccccc.
	rule ccc-> bbb at position 2
.babbbcc.
	rule bbb-> acc at position 2
.baacccc.
	rule ccc-> bbb at position 3
.baabbbc.
	rule bbb-> acc at position 3
.baaaccc.
	rule aaa-> bbc at position 1
.bbbcccc.
	rule bbb-> acc at position 0
.acccccc.
	rule ccc-> bbb at position 1
.abbbccc.
	rule bbb-> acc at position 1
.aaccccc.
	rule ccc-> bbb at position 2
.aabbbcc.
	rule bbb-> acc at position 2
.aaacccc.