Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/zr04new)

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

t(u(x1)) → t(c(d(x1))) (1)
d(f(x1)) → f(d(x1)) (2)
d(g(x1)) → u(g(x1)) (3)
f(u(x1)) → u(f(x1)) (4)
d(n(x1)) → d(x1) (5)
d(o(x1)) → d(x1) (6)
o(u(x1)) → u(x1) (7)

and S is the following TRS.

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) (6)
o(u(x1)) → u(x1) (7)

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
[o(x1)] =
1 2
0 0
· x1 +
15
0
[n(x1)] =
1 0
0 1
· x1 +
0
0
[g(x1)] =
1 0
0 1
· x1 +
3
3
[f(x1)] =
1 0
0 1
· x1 +
12
0
[d(x1)] =
1 11
0 4
· x1 +
0
0
[c(x1)] =
1 1
0 0
· x1 +
0
0
[u(x1)] =
1 4
0 2
· x1 +
0
0
[t(x1)] =
1 6
0 0
· x1 +
9
0
all of the following rules can be deleted.
d(g(x1)) → u(g(x1)) (3)

1.1.1 Rule Removal

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

1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
f(d(x1)) → d(f(x1)) (17)
u(f(x1)) → f(u(x1)) (18)
n(d(x1)) → d(x1) (19)
u(n(x1)) → u(x1) (20)
f(x1) → n(f(x1)) (21)
t(x1) → n(c(t(x1))) (22)
n(c(x1)) → c(n(x1)) (23)
o(c(x1)) → c(o(x1)) (24)
o(c(x1)) → o(x1) (25)
f(c(x1)) → c(f(x1)) (26)
u(c(x1)) → c(u(x1)) (27)
d(c(x1)) → c(d(x1)) (28)

1.1.1.1.1 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)] =
4
· x1 +
3
[c(x1)] =
1
· x1 +
0
[u(x1)] =
8
· x1 +
3
[t(x1)] =
8
· x1 +
3
all of the following rules can be deleted.
f(d(x1)) → d(f(x1)) (17)

1.1.1.1.1.1 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)] =
1
· x1 +
3
[c(x1)] =
1
· x1 +
0
[u(x1)] =
4
· x1 +
1
[t(x1)] =
8
· x1 +
3
all of the following rules can be deleted.
u(f(x1)) → f(u(x1)) (18)

1.1.1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
[o(x1)] =
1 15
0 15
· x1 +
15
15
[n(x1)] =
1 1
0 8
· x1 +
0
0
[f(x1)] =
1 3
0 0
· x1 +
5
0
[d(x1)] =
1 15
0 9
· x1 +
15
1
[c(x1)] =
1 0
0 1
· x1 +
0
0
[u(x1)] =
1 9
0 13
· x1 +
15
0
[t(x1)] =
1 15
0 0
· x1 +
3
0
all of the following rules can be deleted.
n(d(x1)) → d(x1) (19)

1.1.1.1.1.1.1.1 R is empty

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