(1) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite.
Found the self-embedding DerivationStructure:
b c c c c → c c b c c c c b b c b b b b b b b
b c c c c →
c c b c c c c b b c b b b b b b bby OverlapClosure OC 3
b c c c c → c c b c c a c c b b b b b b b
by OverlapClosure OC 3b c c c c → c c b c c b c c b b b b b
by OverlapClosure OC 3b c c c c → c c b c c b b c b c b b b
by OverlapClosure OC 3b c c c c → c c b c c b b c b b c b b c b b b
by OverlapClosure OC 3b c c c c → c c b a c c b b c b b c b b b
by OverlapClosure OC 3b c c c c → c c b a a c c b b c b b b
by OverlapClosure OC 3b c c c c → c c b a a a c c b b b
by OverlapClosure OC 3b c c c c → c c b a a a a c b
by OverlapClosure OC 3b c c c c → c c b a a b c c b
by OverlapClosure OC 2b c c → c c b a a b b
by OverlapClosure OC 3b c c → c c b b c b b
by OverlapClosure OC 3b c c → a c c b b
by OverlapClosure OC 2b c → a a
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
b c → a a
by original rule (OC 1)
b c c → c c b
by OverlapClosure OC 2b c c → c c b b c b b
by OverlapClosure OC 3b c c → a c c b b
by OverlapClosure OC 2b c → a a
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
b c b b →
by OverlapClosure OC 2b c b → a
by OverlapClosure OC 2b c → a a
by original rule (OC 1)
a b →
by original rule (OC 1)
a b →
by original rule (OC 1)
b c → a a
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
b c b b →
by OverlapClosure OC 2b c b → a
by OverlapClosure OC 2b c → a a
by original rule (OC 1)
a b →
by original rule (OC 1)
a b →
by original rule (OC 1)
b c b c → c c b b
by OverlapClosure OC 2b c b → a
by OverlapClosure OC 2b c → a a
by original rule (OC 1)
a b →
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
b c c → a c c b b
by OverlapClosure OC 2b c → a a
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)
a c → c c b b
by original rule (OC 1)