Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-126)

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

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

and S is the following TRS.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(a) = 1 weight(a) = 1
prec(b) = 0 weight(b) = 1
prec(c) = 2 weight(c) = 1
all of the following rules can be deleted.
a(b(c(x1))) b(a(a(x1))) (1)
a(b(b(x1))) b(c(a(x1))) (2)
c(a(a(x1))) c(a(b(x1))) (3)
a(b(a(x1))) a(b(a(x1))) (4)
c(b(c(x1))) a(c(c(x1))) (5)
c(b(a(x1))) a(a(b(x1))) (6)

1.1 R is empty

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