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.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[t(x1)] = 1 · x1
[u(x1)] = 1 · x1
[c(x1)] = 1 · x1
[d(x1)] = 1 · x1
[f(x1)] = 1 · x1
[g(x1)] = 1 · x1
[n(x1)] = 1 · x1
[o(x1)] = 1 + 1 · x1
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 integers
[t(x1)] =
0
0
+
1 2
0 0
· x1
[u(x1)] =
0
0
+
1 0
0 1
· x1
[c(x1)] =
0
0
+
1 0
0 0
· x1
[d(x1)] =
0
0
+
1 2
0 2
· x1
[f(x1)] =
0
0
+
2 0
0 2
· x1
[g(x1)] =
0
2
+
2 0
0 2
· x1
[n(x1)] =
0
0
+
1 0
0 1
· x1
[o(x1)] =
0
0
+
1 0
0 0
· x1
all of the following rules can be deleted.
d(g(x1)) → u(g(x1)) (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[t(x1)] = 1 · x1
[u(x1)] = 1 + 1 · x1
[c(x1)] = 1 · x1
[d(x1)] = 1 · x1
[f(x1)] = 1 · x1
[n(x1)] = 1 · x1
[o(x1)] = 1 · x1
all of the following rules can be deleted.
t(u(x1)) → t(c(d(x1))) (1)

1.1.1.1 Rule Removal

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

1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1.1 R is empty

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