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