Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z104)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c(x1)] = 1 · x1 + 13
[a(x1)] = 1 · x1 + 3
[d(x1)] = 1 · x1 + 21
[b(x1)] = 1 · x1 + 5
all of the following rules can be deleted.
c(x1) a(a(a(a(x1)))) (3)
d(x1) b(b(b(b(x1)))) (4)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c(x1)] = 1 · x1
[a(x1)] = 1 · x1
[d(x1)] = 1 · x1
[b(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
d(b(x1)) c(c(x1)) (2)
b(d(x1)) c(c(x1)) (5)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c(x1)] = 1 · x1
[a(x1)] = 1 · x1 + 1
[d(x1)] = 1 · x1
all of the following rules can be deleted.
c(c(c(a(x1)))) d(d(x1)) (1)
a(c(c(c(x1)))) d(d(x1)) (6)

1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.