Certification Problem

Input (TPDB TRS_Relative/Mixed_relative_TRS/carbridge)

The relative rewrite relation R/S is considered where R is the following TRS

top(left(car(x,y),car(old,z))) → top(right(y,car(old,z))) (1)
top(left(car(x,car(old,y)),z)) → top(right(car(old,y),z)) (2)
top(right(x,car(y,car(old,z)))) → top(left(x,car(old,z))) (3)
top(right(car(old,x),car(y,z))) → top(left(car(old,x),z)) (4)
top(left(bot,car(old,x))) → top(right(bot,car(old,x))) (5)
top(right(car(old,x),bot)) → top(left(car(old,x),bot)) (6)

and S is the following TRS.

top(left(car(x,y),z)) → top(left(y,z)) (7)
top(right(x,car(y,z))) → top(right(x,z)) (8)
bot → car(new,bot) (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
[right(x1, x2)] =
1 1 0
0 0 1
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
1 0 0
0 0 0
[old] =
0 0 0
1 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
[bot] =
0 0 0
0 0 0
1 0 0
[car(x1, x2)] =
1 0 0
0 1 0
0 0 0
· x1 +
1 1 0
0 0 0
0 0 1
· x2 +
0 0 0
0 0 0
0 0 0
[new] =
0 0 0
0 0 0
0 0 0
[left(x1, x2)] =
1 0 0
0 1 1
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
top(right(x,car(y,car(old,z)))) → top(left(x,car(old,z))) (3)
top(right(car(old,x),car(y,z))) → top(left(car(old,x),z)) (4)
top(right(car(old,x),bot)) → top(left(car(old,x),bot)) (6)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[right(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 1 1
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
1 0 0
[old] =
0 0 0
1 0 0
1 0 0
[top(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
[bot] =
0 0 0
0 0 0
0 0 0
[car(x1, x2)] =
1 0 0
0 1 1
1 0 1
· x1 +
1 0 0
0 0 0
0 1 1
· x2 +
0 0 0
0 0 0
0 0 0
[new] =
0 0 0
0 0 0
0 0 0
[left(x1, x2)] =
1 0 1
0 0 1
0 0 0
· x1 +
1 1 1
0 0 1
0 0 1
· x2 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
top(left(car(x,car(old,y)),z)) → top(right(car(old,y),z)) (2)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[right(x1, x2)] =
1 0 0
0 0 0
0 1 0
· x1 +
1 0 0
1 0 0
0 1 0
· x2 +
0 0 0
0 0 0
0 0 0
[old] =
1 0 0
1 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
[bot] =
0 0 0
1 0 0
0 0 0
[car(x1, x2)] =
1 1 0
1 1 1
0 0 0
· x1 +
1 0 0
0 0 1
1 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[new] =
0 0 0
0 0 0
0 0 0
[left(x1, x2)] =
1 0 0
0 0 0
0 0 1
· x1 +
1 0 0
0 0 0
1 0 0
· x2 +
1 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
top(left(car(x,y),car(old,z))) → top(right(y,car(old,z))) (1)
top(left(bot,car(old,x))) → top(right(bot,car(old,x))) (5)

1.1.1.1 R is empty

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