Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/zr08)

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

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

and S is the following TRS.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

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

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[d(x1)] = 1 + 1 · x1
[a(x1)] = 1 + 1 · x1
[c(x1)] = 1 + 1 · x1
[b(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
d(d(d(x1))) → a(c(x1)) (3)

1.1.1.1 R is empty

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