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 matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
[d(x1)] =
1 8
0 1
· x1 +
0
0
[c(x1)] =
1 3
0 1
· x1 +
15
15
[b(x1)] =
1 3
0 0
· x1 +
0
0
[a(x1)] =
1 3
0 0
· x1 +
15
0
all of the following rules can be deleted.
d(c(x1)) → c(x1) (3)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
[d(x1)] =
1 1
0 1
· x1 +
0
0
[c(x1)] =
1 0
0 0
· x1 +
12
0
[b(x1)] =
1 0
0 8
· x1 +
0
1
[a(x1)] =
1 0
0 0
· x1 +
10
0
all of the following rules can be deleted.
d(b(x1)) → b(x1) (2)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 3 with strict dimension 1 over the naturals
[d(x1)] =
1 0 0
0 1 0
0 3 0
· x1 +
0
0
0
[c(x1)] =
1 0 4
0 0 5
0 0 4
· x1 +
7
7
7
[b(x1)] =
1 0 3
0 3 3
0 0 0
· x1 +
0
7
0
[a(x1)] =
1 7 7
0 1 1
0 4 4
· x1 +
7
1
4
all of the following rules can be deleted.
a(b(x1)) → a(x1) (1)

1.1.1.1 R is empty

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