Certification Problem
Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-373)
The rewrite relation of the following TRS is considered.
a(x1) |
→ |
b(x1) |
(1) |
a(c(x1)) |
→ |
x1 |
(2) |
c(b(b(x1))) |
→ |
a(a(a(c(c(x1))))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
No.Proof (by AProVE @ termCOMP 2023)
1 Looping derivation
There is a looping derivation.
c b b b b b b →+ a a a c b b b b b b c c c
The derivation can be derived as follows.
-
c b b →+ a a a c c:
This is an original rule (OC1).
-
a →+ b:
This is an original rule (OC1).
-
c b b →+ b a a c c:
The overlap closure is obtained from the following two overlap closures (OC3).
-
c b b →+ a a a c c
-
a →+ b
-
c b b →+ b b a c c:
The overlap closure is obtained from the following two overlap closures (OC3).
-
c b b →+ b a a c c
-
a →+ b
-
c b b →+ b b b c c:
The overlap closure is obtained from the following two overlap closures (OC3).
-
c b b →+ b b a c c
-
a →+ b
-
a c →+ ε:
This is an original rule (OC1).
-
c b b →+ b b c:
The overlap closure is obtained from the following two overlap closures (OC3).
-
c b b →+ b b a c c
-
a c →+ ε
-
c b b b b →+ b b b c b b c:
The overlap closure is obtained from the following two overlap closures (OC2).
-
c b b →+ b b b c c
-
c b b →+ b b c
-
c b b →+ a b a c c:
The overlap closure is obtained from the following two overlap closures (OC3).
-
c b b →+ a a a c c
-
a →+ b
-
c b b →+ a b b c c:
The overlap closure is obtained from the following two overlap closures (OC3).
-
c b b →+ a b a c c
-
a →+ b
-
c b b b b →+ b b b a b b c c c:
The overlap closure is obtained from the following two overlap closures (OC3).
-
c b b b b →+ b b b c b b c
-
c b b →+ a b b c c
-
c b b b b b b →+ a a a c b b b a b b c c c:
The overlap closure is obtained from the following two overlap closures (OC2).
-
c b b →+ a a a c c
-
c b b b b →+ b b b a b b c c c
-
c b b b b b b →+ a a a c b b b b b b c c c:
The overlap closure is obtained from the following two overlap closures (OC3).
-
c b b b b b b →+ a a a c b b b a b b c c c
-
a →+ b