Certification Problem

Input (TPDB SRS_Relative/Zantema_06_relative/rel04)

The relative rewrite relation R/S is considered where R is the following TRS

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

and S is the following TRS.

a(b(x1)) → d(x1) (4)
d(x1) → a(b(x1)) (5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(c(b(x1))) → d(d(x1)) (6)
b(x1) → c(c(x1)) (2)
a(a(x1)) → a(b(c(a(x1)))) (7)
b(a(x1)) → d(x1) (8)
d(x1) → b(a(x1)) (9)

1.1 Rule Removal

Using the matrix interpretations of dimension 3 with strict dimension 1 over the integers
[a(x1)] =
0
0
0
+
1 0 2
0 0 1
0 0 0
· x1
[c(x1)] =
0
0
0
+
1 0 0
0 0 0
0 2 0
· x1
[b(x1)] =
0
2
0
+
1 0 0
0 0 2
0 0 0
· x1
[d(x1)] =
0
2
0
+
1 0 2
0 0 0
0 0 0
· x1
all of the following rules can be deleted.
a(c(b(x1))) → d(d(x1)) (6)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[b(x1)] =
0
0
+
1 0
0 0
· x1
[c(x1)] =
0
0
+
1 0
0 0
· x1
[a(x1)] =
0
2
+
1 1
0 0
· x1
[d(x1)] =
0
0
+
1 1
0 0
· x1
all of the following rules can be deleted.
a(a(x1)) → a(b(c(a(x1)))) (7)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b(x1)] = 1 + 1 · x1
[c(x1)] = 1 · x1
[a(x1)] = 1 · x1
[d(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
b(x1) → c(c(x1)) (2)

1.1.1.1.1 R is empty

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