Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-173)

The relative rewrite relation R/S is considered where R is the following TRS

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

and S is the following TRS.

c(a(b(x1))) b(c(c(x1))) (3)
a(b(c(x1))) b(b(b(x1))) (4)
b(b(c(x1))) a(b(b(x1))) (5)
a(b(a(x1))) c(a(c(x1))) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[c(x1)] =
1
0
+
1 1
1 0
· x1
[a(x1)] =
2
1
+
1 1
1 0
· x1
[b(x1)] =
0
0
+
1 1
1 0
· x1
all of the following rules can be deleted.
b(a(c(x1))) b(c(b(x1))) (2)
c(a(b(x1))) b(c(c(x1))) (3)
a(b(c(x1))) b(b(b(x1))) (4)
a(b(a(x1))) c(a(c(x1))) (6)

1.1 Rule Removal

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

1.1.1 R is empty

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