Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/dup02)

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

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

and S is the following TRS.

t(t(u(u(x1)))) → t(t(c(c(d(d(x1)))))) (4)
d(d(f(f(x1)))) → f(f(d(d(x1)))) (5)
d(d(g(g(x1)))) → u(u(g(g(x1)))) (6)
f(f(u(u(x1)))) → u(u(f(f(x1)))) (7)
n(n(u(u(x1)))) → u(u(x1)) (8)
f(f(x1)) → f(f(n(n(x1)))) (9)
t(t(x1)) → t(t(c(c(n(n(x1)))))) (10)
c(c(n(n(x1)))) → n(n(c(c(x1)))) (11)
c(c(o(o(x1)))) → o(o(c(c(x1)))) (12)
c(c(o(o(x1)))) → o(o(x1)) (13)
c(c(f(f(x1)))) → f(f(c(c(x1)))) (14)
c(c(u(u(x1)))) → u(u(c(c(x1)))) (15)
c(c(d(d(x1)))) → d(d(c(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(d(o(o(x1)))) → d(d(x1)) (2)
o(o(u(u(x1)))) → u(u(x1)) (3)

1.1 Rule Removal

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

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(t(u(u(x1)))) → t(t(c(c(d(d(x1)))))) (4)

1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
n(n(d(d(x1)))) → d(d(x1)) (17)
f(f(d(d(x1)))) → d(d(f(f(x1)))) (18)
u(u(f(f(x1)))) → f(f(u(u(x1)))) (19)
u(u(n(n(x1)))) → u(u(x1)) (20)
f(f(x1)) → n(n(f(f(x1)))) (21)
t(t(x1)) → n(n(c(c(t(t(x1)))))) (22)
n(n(c(c(x1)))) → c(c(n(n(x1)))) (23)
o(o(c(c(x1)))) → c(c(o(o(x1)))) (24)
o(o(c(c(x1)))) → o(o(x1)) (25)
f(f(c(c(x1)))) → c(c(f(f(x1)))) (26)
u(u(c(c(x1)))) → c(c(u(u(x1)))) (27)
d(d(c(c(x1)))) → c(c(d(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 +
0
[n(x1)] =
1
· x1 +
0
[f(x1)] =
8
· x1 +
2
[d(x1)] =
8
· x1 +
3
[c(x1)] =
1
· x1 +
0
[u(x1)] =
8
· x1 +
2
[t(x1)] =
8
· x1 +
0
all of the following rules can be deleted.
f(f(d(d(x1)))) → d(d(f(f(x1)))) (18)

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 +
0
[n(x1)] =
1
· x1 +
0
[f(x1)] =
8
· x1 +
1
[d(x1)] =
8
· x1 +
3
[c(x1)] =
1
· x1 +
0
[u(x1)] =
8
· x1 +
0
[t(x1)] =
8
· x1 +
0
all of the following rules can be deleted.
u(u(f(f(x1)))) → f(f(u(u(x1)))) (19)

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 1
0 0
· x1 +
4
1
[n(x1)] =
1 5
0 2
· x1 +
0
0
[f(x1)] =
1 0
0 0
· x1 +
7
0
[d(x1)] =
1 1
0 0
· x1 +
7
1
[c(x1)] =
1 0
0 1
· x1 +
0
0
[u(x1)] =
1 0
0 0
· x1 +
3
0
[t(x1)] =
1 8
0 0
· x1 +
3
0
all of the following rules can be deleted.
n(n(d(d(x1)))) → d(d(x1)) (17)

1.1.1.1.1.1.1.1 R is empty

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