Certification Problem

Input (TPDB SRS_Standard/Zantema_06/beans6)

The rewrite relation of the following TRS is considered.

b(a(a(x1))) a(b(c(x1))) (1)
c(a(x1)) a(c(x1)) (2)
c(b(x1)) b(a(x1)) (3)
a(a(b(x1))) d(b(a(x1))) (4)
a(d(x1)) d(a(x1)) (5)
b(d(x1)) a(b(x1)) (6)
a(a(x1)) a(b(a(x1))) (7)

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.
b#(a(a(x1))) a#(b(c(x1))) (8)
b#(a(a(x1))) b#(c(x1)) (9)
b#(a(a(x1))) c#(x1) (10)
c#(a(x1)) a#(c(x1)) (11)
c#(a(x1)) c#(x1) (12)
c#(b(x1)) b#(a(x1)) (13)
c#(b(x1)) a#(x1) (14)
a#(a(b(x1))) b#(a(x1)) (15)
a#(a(b(x1))) a#(x1) (16)
a#(d(x1)) a#(x1) (17)
b#(d(x1)) a#(b(x1)) (18)
b#(d(x1)) b#(x1) (19)
a#(a(x1)) a#(b(a(x1))) (20)
a#(a(x1)) b#(a(x1)) (21)

1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[b#(x1)] = 1 · x1
[a(x1)] = 1 + 1 · x1
[a#(x1)] = 1 · x1
[b(x1)] = 1 · x1
[c(x1)] = 1 + 1 · x1
[c#(x1)] = 1 + 1 · x1
[d(x1)] = 1 + 1 · x1
the pairs
b#(a(a(x1))) a#(b(c(x1))) (8)
b#(a(a(x1))) b#(c(x1)) (9)
b#(a(a(x1))) c#(x1) (10)
c#(a(x1)) a#(c(x1)) (11)
c#(a(x1)) c#(x1) (12)
c#(b(x1)) a#(x1) (14)
a#(a(b(x1))) a#(x1) (16)
a#(d(x1)) a#(x1) (17)
b#(d(x1)) a#(b(x1)) (18)
b#(d(x1)) b#(x1) (19)
could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.