Certification Problem
Input (TPDB SRS_Standard/Bouchare_06/09)
The rewrite relation of the following TRS is considered.
|
b(a(a(x1))) |
→ |
a(x1) |
(1) |
|
a(a(a(x1))) |
→ |
b(b(b(x1))) |
(2) |
|
b(b(x1)) |
→ |
a(b(a(x1))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
No.Proof (by AProVE @ termCOMP 2023)
1 Looping derivation
There is a looping derivation.
b b a a →+ ε b b a a ε
The derivation can be derived as follows.
-
b b →+ a b a:
This is an original rule (OC1).
-
a a a →+ b b b:
This is an original rule (OC1).
-
a a a →+ b a b a:
The overlap closure is obtained from the following two overlap closures (OC2).
-
a a a →+ b b b
-
b b →+ a b a
-
b b a a →+ a b b a b a:
The overlap closure is obtained from the following two overlap closures (OC2).
-
b b →+ a b a
-
a a a →+ b a b a
-
b a a →+ a:
This is an original rule (OC1).
-
b b a →+ a a:
The overlap closure is obtained from the following two overlap closures (OC2).
-
b b a a →+ a a a b a:
The overlap closure is obtained from the following two overlap closures (OC3).
-
b b a a →+ a b b a b a
-
b b a →+ a a
-
b b a a →+ b b b b a:
The overlap closure is obtained from the following two overlap closures (OC3).
-
b b a a →+ a a a b a
-
a a a →+ b b b
-
b b a a →+ b b a b a a:
The overlap closure is obtained from the following two overlap closures (OC3).
-
b b a a →+ b b b b a
-
b b →+ a b a
-
b b a a →+ b b a a:
The overlap closure is obtained from the following two overlap closures (OC2).
-
b b a a →+ b b a b a a
-
b a a →+ a