Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z003)

The rewrite relation of the following TRS is considered.

b(c(a(x1))) → a(b(a(b(c(x1))))) (1)
b(x1) → c(c(x1)) (2)
c(d(x1)) → a(b(c(a(x1)))) (3)
a(a(x1)) → a(c(b(a(x1)))) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(c(b(x1))) → c(b(a(b(a(x1))))) (5)
b(x1) → c(c(x1)) (2)
d(c(x1)) → a(c(b(a(x1)))) (6)
a(a(x1)) → a(b(c(a(x1)))) (7)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[c(x1)] = 0 · x1 + -∞
[d(x1)] = 4 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
[b(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
d(c(x1)) → a(c(b(a(x1)))) (6)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 1
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 0
0 0 1
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
a(c(b(x1))) → c(b(a(b(a(x1))))) (5)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 1 0
1 1 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[b(x1)] =
1 0 0
1 1 0
1 1 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
a(a(x1)) → a(b(c(a(x1)))) (7)

1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1 R is empty

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