Certification Problem

Input (TPDB TRS_Standard/AProVE_04/Liveness6.2)

The rewrite relation of the following TRS is considered.

top(sent(x)) → top(check(rest(x))) (1)
rest(nil) → sent(nil) (2)
rest(cons(x,y)) → sent(y) (3)
check(sent(x)) → sent(check(x)) (4)
check(rest(x)) → rest(check(x)) (5)
check(cons(x,y)) → cons(check(x),y) (6)
check(cons(x,y)) → cons(x,check(y)) (7)
check(cons(x,y)) → cons(x,y) (8)

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
[nil] =
1 0 0
0 0 0
0 0 0
[top(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[check(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[cons(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
[sent(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[rest(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
rest(cons(x,y)) → sent(y) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[nil] =
1 0 0
1 0 0
0 0 0
[top(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[check(x1)] =
1 0 0
0 0 0
1 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[cons(x1, x2)] =
1 0 0
0 0 0
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
1 0 0
[sent(x1)] =
1 0 0
0 1 0
1 1 0
· x1 +
1 0 0
0 0 0
0 0 0
[rest(x1)] =
1 1 0
0 1 0
1 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
top(sent(x)) → top(check(rest(x))) (1)

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(cons) = 2 weight(cons) = 0
prec(nil) = 4 weight(nil) = 1
prec(check) = 3 weight(check) = 4
prec(rest) = 1 weight(rest) = 4
prec(sent) = 0 weight(sent) = 4
all of the following rules can be deleted.
rest(nil) → sent(nil) (2)
check(sent(x)) → sent(check(x)) (4)
check(rest(x)) → rest(check(x)) (5)
check(cons(x,y)) → cons(check(x),y) (6)
check(cons(x,y)) → cons(x,check(y)) (7)
check(cons(x,y)) → cons(x,y) (8)

1.1.1.1 R is empty

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