Certification Problem

Input (TPDB TRS_Standard/Rubio_04/lindau)

The rewrite relation of the following TRS is considered.

c(b(a(X))) → a(a(b(b(c(c(X)))))) (1)
a(X) → e (2)
b(X) → e (3)
c(X) → e (4)

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)] =
1 0 0
0 1 1
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[e] =
0 0 0
1 0 0
0 0 0
[a(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
1 0 0
1 0 0
0 0 0
[c(x1)] =
1 1 0
0 0 0
1 0 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
a(X) → e (2)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(e) = 0 status(e) = [] list-extension(e) = Lex
prec(c) = 2 status(c) = [1] list-extension(c) = Lex
prec(b) = 0 status(b) = [1] list-extension(b) = Lex
and the following Max-polynomial interpretation
[e] = max(0)
[c(x1)] = max(0, 2 + 1 · x1)
[b(x1)] = max(0, 1 + 1 · x1)
all of the following rules can be deleted.
b(X) → e (3)
c(X) → e (4)

1.1.1.1 R is empty

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