Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/zr04)

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

d(n(x1)) → d(x1) (1)
d(o(x1)) → d(x1) (2)
o(u(x1)) → u(x1) (3)

and S is the following TRS.

t(u(x1)) → t(c(d(x1))) (4)
d(f(x1)) → f(d(x1)) (5)
d(g(x1)) → u(g(x1)) (6)
f(u(x1)) → u(f(x1)) (7)
n(u(x1)) → u(x1) (8)
f(x1) → f(n(x1)) (9)
t(x1) → t(c(n(x1))) (10)
c(n(x1)) → n(c(x1)) (11)
c(o(x1)) → o(c(x1)) (12)
c(o(x1)) → o(x1) (13)
c(f(x1)) → f(c(x1)) (14)
c(u(x1)) → u(c(x1)) (15)
c(d(x1)) → d(c(x1)) (16)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[o(x1)] = x1 +
1
[n(x1)] = x1 +
0
[g(x1)] = x1 +
0
[f(x1)] = x1 +
0
[d(x1)] = x1 +
0
[c(x1)] = x1 +
0
[u(x1)] = x1 +
0
[t(x1)] = x1 +
0
all of the following rules can be deleted.
d(o(x1)) → d(x1) (2)
o(u(x1)) → u(x1) (3)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
[o(x1)] =
1 0
0 0
· x1 +
15
0
[n(x1)] =
1 0
0 1
· x1 +
0
0
[g(x1)] =
1 7
0 1
· x1 +
0
3
[f(x1)] =
1 0
0 2
· x1 +
0
0
[d(x1)] =
1 3
0 4
· x1 +
14
0
[c(x1)] =
1 0
0 0
· x1 +
0
0
[u(x1)] =
1 0
0 4
· x1 +
15
0
[t(x1)] =
1 2
0 0
· x1 +
1
0
all of the following rules can be deleted.
t(u(x1)) → t(c(d(x1))) (4)
d(g(x1)) → u(g(x1)) (6)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the naturals
[o(x1)] =
8
· x1 +
3
[n(x1)] =
1
· x1 +
0
[f(x1)] =
8
· x1 +
0
[d(x1)] =
8
· x1 +
0
[c(x1)] =
1
· x1 +
0
[u(x1)] =
8
· x1 +
1
[t(x1)] =
2
· x1 +
3
all of the following rules can be deleted.
f(u(x1)) → u(f(x1)) (7)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the naturals
[o(x1)] =
8
· x1 +
3
[n(x1)] =
1
· x1 +
0
[f(x1)] =
8
· x1 +
3
[d(x1)] =
8
· x1 +
2
[c(x1)] =
1
· x1 +
0
[u(x1)] =
8
· x1 +
3
[t(x1)] =
2
· x1 +
3
all of the following rules can be deleted.
d(f(x1)) → f(d(x1)) (5)

1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
[o(x1)] =
1 1
0 0
· x1 +
15
4
[n(x1)] =
1 0
0 1
· x1 +
0
1
[f(x1)] =
1 0
0 0
· x1 +
1
0
[d(x1)] =
1 1
0 0
· x1 +
7
0
[c(x1)] =
1 0
0 1
· x1 +
0
0
[u(x1)] =
1 0
0 0
· x1 +
8
1
[t(x1)] =
1 0
0 0
· x1 +
12
15
all of the following rules can be deleted.
d(n(x1)) → d(x1) (1)

1.1.1.1.1.1 R is empty

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