Certification Problem

Input (TPDB TRS_Standard/SK90/4.25)

The rewrite relation of the following TRS is considered.

rev(a) a (1)
rev(b) b (2)
rev(++(x,y)) ++(rev(y),rev(x)) (3)
rev(++(x,x)) rev(x) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(++) = 0 weight(++) = 0
prec(b) = 1 weight(b) = 1
prec(rev) = 3 weight(rev) = 0
prec(a) = 2 weight(a) = 1
all of the following rules can be deleted.
rev(a) a (1)
rev(b) b (2)
rev(++(x,y)) ++(rev(y),rev(x)) (3)
rev(++(x,x)) rev(x) (4)

1.1 R is empty

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