Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z112)

The rewrite relation of the following TRS is considered.

a(a(x1)) → b(c(x1)) (1)
b(b(x1)) → c(d(x1)) (2)
b(x1) → a(x1) (3)
c(c(x1)) → d(f(x1)) (4)
d(d(x1)) → f(f(f(x1))) (5)
d(x1) → b(x1) (6)
f(f(x1)) → g(a(x1)) (7)
g(g(x1)) → a(x1) (8)

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(a(x1)) → c(b(x1)) (9)
b(b(x1)) → d(c(x1)) (10)
b(x1) → a(x1) (3)
c(c(x1)) → f(d(x1)) (11)
d(d(x1)) → f(f(f(x1))) (5)
d(x1) → b(x1) (6)
f(f(x1)) → a(g(x1)) (12)
g(g(x1)) → a(x1) (8)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[d(x1)] =
1 0 1
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[g(x1)] =
1 1 0
0 0 1
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[a(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
g(g(x1)) → a(x1) (8)

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(g) = 7 weight(g) = 0
prec(f) = 1 weight(f) = 2
prec(d) = 5 weight(d) = 4
prec(b) = 4 weight(b) = 4
prec(c) = 2 weight(c) = 3
prec(a) = 0 weight(a) = 4
all of the following rules can be deleted.
a(a(x1)) → c(b(x1)) (9)
b(b(x1)) → d(c(x1)) (10)
b(x1) → a(x1) (3)
c(c(x1)) → f(d(x1)) (11)
d(d(x1)) → f(f(f(x1))) (5)
d(x1) → b(x1) (6)
f(f(x1)) → a(g(x1)) (12)

1.1.1.1 R is empty

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