Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/zr12)

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

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

and S is the following TRS.

a(x1) → d(a(x1)) (4)
c(x1) → c(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
b(a(x1)) → a(x1) (6)
b(d(x1)) → b(x1) (7)
c(d(x1)) → c(x1) (8)
a(x1) → a(d(x1)) (9)
c(x1) → b(c(x1)) (10)

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[c(x1)] =
0
0
+
1 1
0 0
· x1
[d(x1)] =
0
1
+
1 0
0 1
· x1
[a(x1)] =
1
2
+
1 0
0 0
· x1
[b(x1)] =
0
0
+
1 0
0 0
· x1
all of the following rules can be deleted.
c(d(x1)) → c(x1) (8)

1.1.1.1.1 R is empty

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