Certification Problem

Input (TPDB SRS_Relative/Zantema_06_relative/rel08)

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

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

and S is the following TRS.

b(a(x1)) → a(b(x1)) (5)
a(b(x1)) → b(a(x1)) (6)

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
[a(x1)] =
0
2
+
1 0
0 1
· x1
[b(x1)] =
0
1
+
1 0
0 1
· x1
[c(x1)] =
0
0
+
1 0
0 0
· x1
[d(x1)] =
0
0
+
1 1
0 2
· x1
all of the following rules can be deleted.
d(b(b(b(x1)))) → a(d(a(x1))) (4)

1.1 Rule Removal

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

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 1
· x1
[a(x1)] =
0
0
+
2 1
0 1
· x1
[c(x1)] =
0
1
+
1 1
0 0
· x1
all of the following rules can be deleted.
a(c(x1)) → b(b(c(a(x1)))) (3)

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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