Certification Problem

Input (TPDB TRS_Standard/AProVE_04/Liveness_WRS)

The rewrite relation of the following TRS is considered.

active(f(x)) → mark(f(f(x))) (1)
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))),x))) (2)
mat(f(x),f(y)) → f(mat(x,y)) (3)
chk(no(c)) → active(c) (4)
mat(f(x),c) → no(c) (5)
f(active(x)) → active(f(x)) (6)
f(no(x)) → no(f(x)) (7)
f(mark(x)) → mark(f(x)) (8)
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))),x))) (9)

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
[chk(x1)] =
1 1 0
1 0 1
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[c] =
0 0 0
0 0 0
0 0 0
[active(x1)] =
1 1 0
1 0 0
1 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[no(x1)] =
1 1 0
1 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[X] =
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 1 0
1 0 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[y] =
0 0 0
1 0 0
0 0 0
[tp(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[mat(x1, x2)] =
1 0 0
0 1 0
1 0 0
· x1 +
1 0 0
1 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[mark(x1)] =
1 0 0
0 1 0
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
mat(f(x),f(y)) → f(mat(x,y)) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[chk(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[c] =
0 0 0
1 0 0
0 0 0
[active(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[no(x1)] =
1 1 1
0 1 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[X] =
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[tp(x1)] =
1 1 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[mat(x1, x2)] =
1 0 0
0 0 0
1 0 0
· x1 +
1 1 0
0 1 0
0 1 0
· x2 +
0 0 0
1 0 0
0 0 0
[mark(x1)] =
1 1 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
chk(no(c)) → active(c) (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[chk(x1)] =
1 1 0
1 1 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[c] =
0 0 0
0 0 0
1 0 0
[active(x1)] =
1 0 1
0 0 0
1 0 0
· x1 +
1 0 0
1 0 0
0 0 0
[no(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
1 0 0
[X] =
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
0 0 1
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[tp(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[mat(x1, x2)] =
1 0 0
0 0 1
0 1 1
· x1 +
1 0 1
0 0 0
0 0 0
· x2 +
0 0 0
1 0 0
1 0 0
[mark(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
1 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))),x))) (2)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(tp) = 0 weight(tp) = 4
prec(c) = 4 weight(c) = 2
prec(mat) = 11 weight(mat) = 0
prec(X) = 2 weight(X) = 4
prec(chk) = 9 weight(chk) = 1
prec(no) = 8 weight(no) = 1
prec(mark) = 1 weight(mark) = 6
prec(active) = 12 weight(active) = 6
prec(f) = 13 weight(f) = 0
all of the following rules can be deleted.
active(f(x)) → mark(f(f(x))) (1)
mat(f(x),c) → no(c) (5)
f(active(x)) → active(f(x)) (6)
f(no(x)) → no(f(x)) (7)
f(mark(x)) → mark(f(x)) (8)
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))),x))) (9)

1.1.1.1.1 R is empty

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