/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 0 2 0 0 ,
  2 2 2 ⟶ 1 }

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 0 2 0 1 ,
  2 2 2 ⟶ 1 }

Loop of length 32 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.aaaaaa
	rule ba-> aacab at position 0
.aacab.aaaaaa
	rule a->  at position 3
.aacb.aaaaaa
	rule ba-> aacab at position 3
.aacaacab.aaaaa
	rule a->  at position 3
.aacacab.aaaaa
	rule a->  at position 3
.aaccab.aaaaa
	rule a->  at position 4
.aaccb.aaaaa
	rule ba-> aacab at position 4
.aaccaacab.aaaa
	rule a->  at position 4
.aaccacab.aaaa
	rule a->  at position 4
.aacccab.aaaa
	rule ccc-> b at position 2
.aabab.aaaa
	rule ba-> aacab at position 4
.aabaaacab.aaa
	rule a->  at position 7
.aabaaacb.aaa
	rule ba-> aacab at position 7
.aabaaacaacab.aa
	rule a->  at position 7
.aabaaacacab.aa
	rule a->  at position 7
.aabaaaccab.aa
	rule a->  at position 8
.aabaaaccb.aa
	rule ba-> aacab at position 8
.aabaaaccaacab.a
	rule a->  at position 8
.aabaaaccacab.a
	rule a->  at position 8
.aabaaacccab.a
	rule ccc-> b at position 6
.aabaaabab.a
	rule ba-> aacab at position 6
.aabaaaaacabb.a
	rule a->  at position 9
.aabaaaaacbb.a
	rule ba-> aacab at position 10
.aabaaaaacbaacab.
	rule ba-> aacab at position 9
.aabaaaaacaacabacab.
	rule a->  at position 9
.aabaaaaacacabacab.
	rule a->  at position 9
.aabaaaaaccabacab.
	rule a->  at position 10
.aabaaaaaccbacab.
	rule ba-> aacab at position 10
.aabaaaaaccaacabcab.
	rule a->  at position 10
.aabaaaaaccacabcab.
	rule a->  at position 10
.aabaaaaacccabcab.
	rule ccc-> b at position 8
.aabaaaaababcab.
	rule ba-> aacab at position 8
.aabaaaaaaacabbcab.