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The given Problem was:
a ->
a b ->
b b -> c
c a -> a a b c
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A self-embedding rewrite structure for the given SRS is:
c a a a a a --(15)-> a c a a a a a b c b c (#140963)
OC+OC at 2
OC: c a a --(6)-> a c c (#140)
OC+OC at 1
OC: c a a --(5)-> a b b c (#65)
OC+OC at 2
OC: c a a --(4)-> a b a b c (#45)
OC+OC at 2
OC: c a --(2)-> a b c (#5)
OC+OC at 0
OC: c a --(1)-> a a b c (#4)
Rule
OC: a --(1)-> (#1)
Rule
OC: c a --(2)-> a b c (#5)
OC+OC at 0
OC: c a --(1)-> a a b c (#4)
Rule
OC: a --(1)-> (#1)
Rule
OC: a --(1)-> (#1)
Rule
OC: b b --(1)-> c (#3)
Rule
OC: c a a a --(9)-> a a a a a b c b c (#7952)
OC+OC at 3
OC: c a a a --(8)-> a a a c a b c (#2512)
OC+OC at 2
OC: c a a a --(6)-> a a c a a b c (#501)
OC+OC at 3
OC: c a a --(5)-> a a c c (#141)
OC+OC at 2
OC: c a a --(4)-> a a b b c (#63)
OC+OC at 3
OC: c a a --(3)-> a a b a b c (#47)
OC+OC at 3
OC: c a --(1)-> a a b c (#4)
Rule
OC: c a --(2)-> a b c (#5)
OC+OC at 0
OC: c a --(1)-> a a b c (#4)
Rule
OC: a --(1)-> (#1)
Rule
OC: a --(1)-> (#1)
Rule
OC: b b --(1)-> c (#3)
Rule
OC: c a --(1)-> a a b c (#4)
Rule
OC: c a --(2)-> a c (#8)
OC+OC at 1
OC: c a --(1)-> a a b c (#4)
Rule
OC: a b --(1)-> (#2)
Rule
OC: c a --(1)-> a a b c (#4)
Rule
Hence, the SRS is non-terminating.