Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/zr02new)

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

p(s(x1)) → s(x1) (1)
n(s(x1)) → s(x1) (2)
o(s(x1)) → s(x1) (3)
o(n(x1)) → n(o(x1)) (4)
o(m(x1)) → n(o(x1)) (5)
n(o(p(x1))) → o(n(x1)) (6)

and S is the following TRS.

t(x1) → t(c(n(x1))) (7)
p(n(x1)) → m(p(x1)) (8)
p(m(x1)) → m(p(x1)) (9)
n(p(x1)) → p(n(x1)) (10)
c(p(x1)) → p(c(x1)) (11)
c(m(x1)) → m(c(x1)) (12)
c(n(x1)) → n(c(x1)) (13)
c(o(x1)) → o(c(x1)) (14)
c(o(x1)) → o(x1) (15)

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
[m(x1)] = x1 +
0
[c(x1)] = x1 +
0
[t(x1)] = x1 +
0
[s(x1)] = x1 +
0
[p(x1)] = x1 +
1
all of the following rules can be deleted.
p(s(x1)) → s(x1) (1)
o(s(x1)) → s(x1) (3)
n(o(p(x1))) → o(n(x1)) (6)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
[o(x1)] =
1 11
0 0
· x1 +
12
0
[n(x1)] =
1 15
0 1
· x1 +
0
0
[m(x1)] =
1 0
0 1
· x1 +
0
0
[c(x1)] =
1 0
0 0
· x1 +
0
0
[t(x1)] =
1 15
0 0
· x1 +
14
15
[s(x1)] =
1 6
0 11
· x1 +
15
8
[p(x1)] =
1 0
0 1
· x1 +
15
0
all of the following rules can be deleted.
n(s(x1)) → s(x1) (2)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
[o(x1)] =
1 10
0 1
· x1 +
11
0
[n(x1)] =
1 0
0 15
· x1 +
0
5
[m(x1)] =
1 0
0 15
· x1 +
0
5
[c(x1)] =
1 0
0 1
· x1 +
0
0
[t(x1)] =
1 0
0 0
· x1 +
15
15
[p(x1)] =
1 0
0 1
· x1 +
7
0
all of the following rules can be deleted.
o(n(x1)) → n(o(x1)) (4)
o(m(x1)) → n(o(x1)) (5)

1.1.1.1 R is empty

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