Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/ExProp7_Luc06_GM)

The rewrite relation of the following TRS is considered.

a__f(0) → cons(0,f(s(0))) (1)
a__f(s(0)) → a__f(a__p(s(0))) (2)
a__p(s(X)) → mark(X) (3)
mark(f(X)) → a__f(mark(X)) (4)
mark(p(X)) → a__p(mark(X)) (5)
mark(0) → 0 (6)
mark(cons(X1,X2)) → cons(mark(X1),X2) (7)
mark(s(X)) → s(mark(X)) (8)
a__f(X) → f(X) (9)
a__p(X) → p(X) (10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[cons(x1, x2)] =
1 0 1
0 0 1
0 1 0
· x1 +
1 0 0
1 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[a__f(x1)] =
1 1 0
0 1 0
0 1 1
· x1 +
0 0 0
0 0 0
1 0 0
[f(x1)] =
1 0 0
0 1 0
0 1 1
· x1 +
0 0 0
0 0 0
1 0 0
[a__p(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[p(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[mark(x1)] =
1 1 1
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 1 1
0 1 1
0 1 1
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
a__f(s(0)) → a__f(a__p(s(0))) (2)
mark(f(X)) → a__f(mark(X)) (4)
mark(s(X)) → s(mark(X)) (8)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(p) = 1 weight(p) = 1
prec(mark) = 7 weight(mark) = 1
prec(a__p) = 3 weight(a__p) = 1
prec(cons) = 4 weight(cons) = 0
prec(f) = 2 weight(f) = 1
prec(s) = 6 weight(s) = 1
prec(a__f) = 0 weight(a__f) = 7
prec(0) = 5 weight(0) = 4
all of the following rules can be deleted.
a__f(0) → cons(0,f(s(0))) (1)
a__p(s(X)) → mark(X) (3)
mark(p(X)) → a__p(mark(X)) (5)
mark(0) → 0 (6)
mark(cons(X1,X2)) → cons(mark(X1),X2) (7)
a__f(X) → f(X) (9)
a__p(X) → p(X) (10)

1.1.1 R is empty

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