Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/9)

The rewrite relation of the following TRS is considered.

a(x,y) → b(x,b(0,c(y))) (1)
c(b(y,c(x))) → c(c(b(a(0,0),y))) (2)
b(y,0) → y (3)

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
[0] =
0 0 0
0 0 0
0 0 0
[b(x1, x2)] =
1 0 1
1 1 0
0 1 1
· x1 +
1 0 0
0 0 0
0 1 0
· x2 +
0 0 0
0 0 0
0 0 0
[a(x1, x2)] =
1 0 1
1 1 0
0 1 1
· x1 +
1 1 1
0 1 0
1 0 1
· x2 +
1 0 0
0 0 0
0 0 0
[c(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
a(x,y) → b(x,b(0,c(y))) (1)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[0] =
0 0 0
0 0 0
1 0 0
[b(x1, x2)] =
1 0 1
1 1 0
1 0 1
· x1 +
1 0 0
1 1 1
1 0 0
· x2 +
1 0 0
0 0 0
1 0 0
[a(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 1 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
b(y,0) → y (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[0] =
0 0 0
0 0 0
0 0 0
[b(x1, x2)] =
1 1 1
1 1 1
0 0 0
· x1 +
1 1 1
0 0 0
0 0 1
· x2 +
0 0 0
1 0 0
1 0 0
[a(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 0 0
1 1 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
c(b(y,c(x))) → c(c(b(a(0,0),y))) (2)

1.1.1.1 R is empty

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