Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z069)

The rewrite relation of the following TRS is considered.

a(b(x1)) C(x1) (1)
b(c(x1)) A(x1) (2)
c(a(x1)) B(x1) (3)
A(C(x1)) b(x1) (4)
C(B(x1)) a(x1) (5)
B(A(x1)) c(x1) (6)
a(a(a(a(x1)))) A(A(A(x1))) (7)
A(A(A(A(x1)))) a(a(a(x1))) (8)
b(b(b(b(x1)))) B(B(B(x1))) (9)
B(B(B(B(x1)))) b(b(b(x1))) (10)
c(c(c(c(x1)))) C(C(C(x1))) (11)
C(C(C(C(x1)))) c(c(c(x1))) (12)
B(a(a(a(x1)))) c(A(A(A(x1)))) (13)
A(A(A(b(x1)))) a(a(a(C(x1)))) (14)
C(b(b(b(x1)))) a(B(B(B(x1)))) (15)
B(B(B(c(x1)))) b(b(b(A(x1)))) (16)
A(c(c(c(x1)))) b(C(C(C(x1)))) (17)
C(C(C(a(x1)))) c(c(c(B(x1)))) (18)
a(A(x1)) x1 (19)
A(a(x1)) x1 (20)
b(B(x1)) x1 (21)
B(b(x1)) x1 (22)
c(C(x1)) x1 (23)
C(c(x1)) x1 (24)

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
[a(x1)] = 1 · x1 + 1
[b(x1)] = 1 · x1 + 1
[C(x1)] = 1 · x1 + 1
[c(x1)] = 1 · x1 + 1
[A(x1)] = 1 · x1 + 1
[B(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
a(b(x1)) C(x1) (1)
b(c(x1)) A(x1) (2)
c(a(x1)) B(x1) (3)
A(C(x1)) b(x1) (4)
C(B(x1)) a(x1) (5)
B(A(x1)) c(x1) (6)
a(a(a(a(x1)))) A(A(A(x1))) (7)
A(A(A(A(x1)))) a(a(a(x1))) (8)
b(b(b(b(x1)))) B(B(B(x1))) (9)
B(B(B(B(x1)))) b(b(b(x1))) (10)
c(c(c(c(x1)))) C(C(C(x1))) (11)
C(C(C(C(x1)))) c(c(c(x1))) (12)
a(A(x1)) x1 (19)
A(a(x1)) x1 (20)
b(B(x1)) x1 (21)
B(b(x1)) x1 (22)
c(C(x1)) x1 (23)
C(c(x1)) x1 (24)

1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
B#(a(a(a(x1)))) A#(A(A(x1))) (25)
B#(a(a(a(x1)))) A#(A(x1)) (26)
B#(a(a(a(x1)))) A#(x1) (27)
A#(A(A(b(x1)))) C#(x1) (28)
C#(b(b(b(x1)))) B#(B(B(x1))) (29)
C#(b(b(b(x1)))) B#(B(x1)) (30)
C#(b(b(b(x1)))) B#(x1) (31)
B#(B(B(c(x1)))) A#(x1) (32)
A#(c(c(c(x1)))) C#(C(C(x1))) (33)
A#(c(c(c(x1)))) C#(C(x1)) (34)
A#(c(c(c(x1)))) C#(x1) (35)
C#(C(C(a(x1)))) B#(x1) (36)

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[B#(x1)] = 1 + 1 · x1
[a(x1)] = 1 + 1 · x1
[A#(x1)] = 1 + 1 · x1
[A(x1)] = 1 + 1 · x1
[b(x1)] = 1 + 1 · x1
[C#(x1)] = 1 + 1 · x1
[B(x1)] = 1 + 1 · x1
[c(x1)] = 1 + 1 · x1
[C(x1)] = 1 + 1 · x1
the pairs
B#(a(a(a(x1)))) A#(A(A(x1))) (25)
B#(a(a(a(x1)))) A#(A(x1)) (26)
B#(a(a(a(x1)))) A#(x1) (27)
A#(A(A(b(x1)))) C#(x1) (28)
C#(b(b(b(x1)))) B#(B(B(x1))) (29)
C#(b(b(b(x1)))) B#(B(x1)) (30)
C#(b(b(b(x1)))) B#(x1) (31)
B#(B(B(c(x1)))) A#(x1) (32)
A#(c(c(c(x1)))) C#(C(C(x1))) (33)
A#(c(c(c(x1)))) C#(C(x1)) (34)
A#(c(c(c(x1)))) C#(x1) (35)
C#(C(C(a(x1)))) B#(x1) (36)
could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.