Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z004)

The rewrite relation of the following TRS is considered.

a(b(a(a(x1)))) → c(b(a(b(a(x1))))) (1)
a(c(b(x1))) → a(a(b(c(b(a(x1)))))) (2)

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(b(a(x1)))) → a(b(a(b(c(x1))))) (3)
b(c(a(x1))) → a(b(c(b(a(a(x1)))))) (4)

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 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 1 0
0 1 1
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[c(x1)] =
1 1 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
a(a(b(a(x1)))) → a(b(a(b(c(x1))))) (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0 1
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[c(x1)] =
1 0 1
1 0 1
0 0 1
· x1 +
0 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
b(c(a(x1))) → a(b(c(b(a(a(x1)))))) (4)

1.1.1.1 R is empty

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