Certification Problem

Input (TPDB SRS_Standard/Zantema_06/while2)

The rewrite relation of the following TRS is considered.

a(b(x1)) → d(x1) (1)
b(a(x1)) → a(b(x1)) (2)
d(c(x1)) → f(a(b(b(c(x1))))) (3)
d(f(x1)) → f(a(b(x1))) (4)
a(f(x1)) → a(x1) (5)

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
b(a(x1)) → d(x1) (6)
a(b(x1)) → b(a(x1)) (7)
c(d(x1)) → c(b(b(a(f(x1))))) (8)
f(d(x1)) → b(a(f(x1))) (9)
f(a(x1)) → a(x1) (10)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0 0
1 1 0
0 1 0
· x1 +
1 0 0
1 0 0
0 0 0
[a(x1)] =
1 0 0
1 0 0
1 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[c(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
0 0 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[d(x1)] =
1 0 0
1 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
f(a(x1)) → a(x1) (10)

1.1.1 Rule Removal

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

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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