/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 2-rule system
{ 0 0 0 ⟶ 1 ,
  1 2 ⟶ 2 2 0 0 0 0 }

The system was reversed.

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

Loop of length 16 starting with a string of length 5
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cb.bbb
	rule cb-> aaaacc at position 0
.aaaacc.bbb
	rule cb-> aaaacc at position 5
.aaaacaaaacc.bb
	rule aaa-> b at position 5
.aaaacbacc.bb
	rule cb-> aaaacc at position 8
.aaaacbacaaaacc.b
	rule aaa-> b at position 8
.aaaacbacbacc.b
	rule cb-> aaaacc at position 7
.aaaacbaaaaaccacc.b
	rule aaa-> b at position 6
.aaaacbbaaccacc.b
	rule cb-> aaaacc at position 13
.aaaacbbaaccacaaaacc.
	rule aaa-> b at position 13
.aaaacbbaaccacbacc.
	rule cb-> aaaacc at position 12
.aaaacbbaaccaaaaaccacc.
	rule aaa-> b at position 11
.aaaacbbaaccbaaccacc.
	rule cb-> aaaacc at position 10
.aaaacbbaacaaaaccaaccacc.
	rule aaa-> b at position 10
.aaaacbbaacbaccaaccacc.
	rule cb-> aaaacc at position 9
.aaaacbbaaaaaaccaccaaccacc.
	rule aaa-> b at position 7
.aaaacbbbaaaccaccaaccacc.
	rule aaa-> b at position 8
.aaaacbbbbccaccaaccacc.