Certification Problem

Input (TPDB TRS_Standard/SK90/4.02)

The rewrite relation of the following TRS is considered.

*(x,1) x (1)
*(1,y) y (2)
*(i(x),x) 1 (3)
*(x,i(x)) 1 (4)
*(x,*(y,z)) *(*(x,y),z) (5)
i(1) 1 (6)
*(*(x,y),i(y)) x (7)
*(*(x,i(y)),y) x (8)
i(i(x)) x (9)
i(*(x,y)) *(i(y),i(x)) (10)
k(x,1) 1 (11)
k(x,x) 1 (12)
*(k(x,y),k(y,x)) 1 (13)
*(*(i(x),k(y,z)),x) k(*(*(i(x),y),x),*(*(i(x),z),x)) (14)
k(*(x,i(y)),*(y,i(x))) 1 (15)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the
prec(*) = 1 stat(*) = lex
prec(1) = 0 stat(1) = lex
prec(i) = 2 stat(i) = lex
prec(k) = 0 stat(k) = lex

π(*) = [2,1]
π(1) = []
π(i) = [1]
π(k) = [1,2]

all of the following rules can be deleted.
*(x,1) x (1)
*(1,y) y (2)
*(i(x),x) 1 (3)
*(x,i(x)) 1 (4)
*(x,*(y,z)) *(*(x,y),z) (5)
i(1) 1 (6)
*(*(x,y),i(y)) x (7)
*(*(x,i(y)),y) x (8)
i(i(x)) x (9)
i(*(x,y)) *(i(y),i(x)) (10)
k(x,1) 1 (11)
k(x,x) 1 (12)
*(k(x,y),k(y,x)) 1 (13)
*(*(i(x),k(y,z)),x) k(*(*(i(x),y),x),*(*(i(x),z),x)) (14)
k(*(x,i(y)),*(y,i(x))) 1 (15)

1.1 R is empty

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