/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, b ↦ 1, c ↦ 2 },
it remains to prove termination of the 4-rule system
{ 0 ⟶ ,
  0 0 0 ⟶ 1 ,
  1 2 ⟶ 2 2 0 1 ,
  2 ⟶ }

Loop of length 13 starting with a string of length 7
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.bc.ccccc
	rule bc-> ccab at position 0
.ccab.ccccc
	rule bc-> ccab at position 3
.ccaccab.cccc
	rule c->  at position 3
.ccacab.cccc
	rule c->  at position 3
.ccaab.cccc
	rule bc-> ccab at position 4
.ccaaccab.ccc
	rule c->  at position 4
.ccaacab.ccc
	rule c->  at position 4
.ccaaab.ccc
	rule aaa-> b at position 2
.ccbb.ccc
	rule bc-> ccab at position 3
.ccbccab.cc
	rule a->  at position 5
.ccbccb.cc
	rule bc-> ccab at position 5
.ccbccccab.c
	rule a->  at position 7
.ccbccccb.c
	rule bc-> ccab at position 7
.ccbccccccab.