Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/8)

The rewrite relation of the following TRS is considered.

b(b(0,y),x) → y (1)
c(c(c(y))) → c(c(a(a(c(b(0,y)),0),0))) (2)
a(y,0) → b(y,0) (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
[b(x1, x2)] =
1 1 0
0 0 1
0 0 1
· x1 +
1 0 0
0 0 0
0 1 1
· x2 +
0 0 0
0 0 0
0 0 0
[a(x1, x2)] =
1 1 0
0 0 1
0 0 1
· x1 +
1 0 0
0 0 0
0 0 1
· x2 +
0 0 0
0 0 0
0 0 0
[0] =
0 0 0
0 0 0
1 0 0
[c(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.
b(b(0,y),x) → y (1)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(a) = 1 weight(a) = 0
prec(b) = 0 weight(b) = 0
prec(0) = 3 weight(0) = 1
all of the following rules can be deleted.
a(y,0) → b(y,0) (3)

1.1.1.1 R is empty

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