Certification Problem

Input (TPDB TRS_Standard/SK90/4.32)

The rewrite relation of the following TRS is considered.

a(b(x)) → b(a(a(x))) (1)
b(c(x)) → c(b(b(x))) (2)
c(a(x)) → a(c(c(x))) (3)
u(a(x)) → x (4)
v(b(x)) → x (5)
w(c(x)) → x (6)
a(u(x)) → x (7)
b(v(x)) → x (8)
c(w(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 the arctic semiring over the integers
[v(x1)] = 5 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
[u(x1)] = 5 · x1 + -∞
[w(x1)] = 1 · x1 + -∞
[b(x1)] = 0 · x1 + -∞
[c(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
u(a(x)) → x (4)
v(b(x)) → x (5)
w(c(x)) → x (6)
a(u(x)) → x (7)
b(v(x)) → x (8)
c(w(x)) → x (9)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 1 1
0 0 1
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
[b(x1)] =
1 1 0
0 1 1
0 1 1
· x1 +
0 0 0
1 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.
a(b(x)) → b(a(a(x))) (1)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(c) = 1 weight(c) = 0
prec(a) = 0 weight(a) = 2
all of the following rules can be deleted.
c(a(x)) → a(c(c(x))) (3)

1.1.1.1.1 R is empty

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