Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z011)

The rewrite relation of the following TRS is considered.

a(b(x1)) b(b(a(x1))) (1)
b(c(x1)) c(b(b(x1))) (2)
c(a(x1)) a(c(c(x1))) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(b(x1)) b#(b(a(x1))) (4)
a#(b(x1)) b#(a(x1)) (5)
a#(b(x1)) a#(x1) (6)
b#(c(x1)) c#(b(b(x1))) (7)
b#(c(x1)) b#(b(x1)) (8)
b#(c(x1)) b#(x1) (9)
c#(a(x1)) a#(c(c(x1))) (10)
c#(a(x1)) c#(c(x1)) (11)
c#(a(x1)) c#(x1) (12)

1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 0
[b(x1)] = 1 · x1
[b#(x1)] = 1 · x1
[a(x1)] = 0
[c(x1)] = 1 + 1 · x1
[c#(x1)] = 0
the pairs
b#(c(x1)) c#(b(b(x1))) (7)
b#(c(x1)) b#(b(x1)) (8)
b#(c(x1)) b#(x1) (9)
could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.