Certification Problem
Input (TPDB SRS_Standard/Secret_07_SRS/num-539)
The rewrite relation of the following TRS is considered.
|
a(a(b(x1))) |
→ |
c(c(a(a(a(x1))))) |
(1) |
|
a(c(x1)) |
→ |
b(a(x1)) |
(2) |
Property / Task
Prove or disprove termination.Answer / Result
No.Proof (by AProVE @ termCOMP 2023)
1 Looping derivation
There is a looping derivation.
a a b b a b b →+ c c b b a a b b a b b a a a a
The derivation can be derived as follows.
-
a a b →+ c c a a a:
This is an original rule (OC1).
-
a a b b →+ c c a c c a a a:
The overlap closure is obtained from the following two overlap closures (OC2).
-
a a b →+ c c a a a
-
a a b →+ c c a a a
-
a c →+ b a:
This is an original rule (OC1).
-
a a b b →+ c c b a c a a a:
The overlap closure is obtained from the following two overlap closures (OC3).
-
a a b b →+ c c a c c a a a
-
a c →+ b a
-
a a b b →+ c c b b a a a a:
The overlap closure is obtained from the following two overlap closures (OC3).
-
a a b b →+ c c b a c a a a
-
a c →+ b a
-
a a b b a b →+ c c b b a a a c c a a a:
The overlap closure is obtained from the following two overlap closures (OC2).
-
a a b b →+ c c b b a a a a
-
a a b →+ c c a a a
-
a a b b a b →+ c c b b a a b a c a a a:
The overlap closure is obtained from the following two overlap closures (OC3).
-
a a b b a b →+ c c b b a a a c c a a a
-
a c →+ b a
-
a a b b a b →+ c c b b a a b b a a a a:
The overlap closure is obtained from the following two overlap closures (OC3).
-
a a b b a b →+ c c b b a a b a c a a a
-
a c →+ b a
-
a a b b a b b →+ c c b b a a b b a a c c a a a:
The overlap closure is obtained from the following two overlap closures (OC2).
-
a a b b a b →+ c c b b a a b b a a a a
-
a a b →+ c c a a a
-
a a b b a b b →+ c c b b a a b b a b a c a a a:
The overlap closure is obtained from the following two overlap closures (OC3).
-
a a b b a b b →+ c c b b a a b b a a c c a a a
-
a c →+ b a
-
a a b b a b b →+ c c b b a a b b a b b a a a a:
The overlap closure is obtained from the following two overlap closures (OC3).
-
a a b b a b b →+ c c b b a a b b a b a c a a a
-
a c →+ b a