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.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

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)
o(o(d(d(x1)))) → d(d(x1)) (18)
u(u(o(o(x1)))) → u(u(x1)) (19)
g(g(d(d(x1)))) → g(g(u(u(x1)))) (20)
o(o(c(c(x1)))) → c(c(o(o(x1)))) (21)
n(n(c(c(x1)))) → c(c(n(n(x1)))) (22)
u(u(n(n(x1)))) → u(u(x1)) (23)
f(f(x1)) → n(n(f(f(x1)))) (24)
u(u(f(f(x1)))) → f(f(u(u(x1)))) (25)
d(d(c(c(x1)))) → c(c(d(d(x1)))) (26)
o(o(c(c(x1)))) → o(o(x1)) (27)
f(f(d(d(x1)))) → d(d(f(f(x1)))) (28)
f(f(c(c(x1)))) → c(c(f(f(x1)))) (29)
u(u(t(t(x1)))) → d(d(c(c(t(t(x1)))))) (30)
t(t(x1)) → n(n(c(c(t(t(x1)))))) (31)
u(u(c(c(x1)))) → c(c(u(u(x1)))) (32)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[n(x1)] = 1 · x1
[d(x1)] = 1 + 1 · x1
[o(x1)] = 1 + 1 · x1
[u(x1)] = 1 + 1 · x1
[g(x1)] = 1 · x1
[c(x1)] = 1 · x1
[f(x1)] = 1 · x1
[t(x1)] = 1 · x1
all of the following rules can be deleted.
o(o(d(d(x1)))) → d(d(x1)) (18)
u(u(o(o(x1)))) → u(u(x1)) (19)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[n(x1)] =
0
0
+
1 0
0 1
· x1
[d(x1)] =
0
2
+
1 0
0 1
· x1
[g(x1)] =
0
0
+
1 2
0 0
· x1
[u(x1)] =
0
0
+
1 0
0 1
· x1
[o(x1)] =
0
0
+
1 0
0 0
· x1
[c(x1)] =
0
0
+
1 0
0 0
· x1
[f(x1)] =
0
0
+
1 0
0 1
· x1
[t(x1)] =
0
2
+
1 0
0 1
· x1
all of the following rules can be deleted.
g(g(d(d(x1)))) → g(g(u(u(x1)))) (20)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[n(x1)] = 1 · x1
[d(x1)] = 1 · x1
[o(x1)] = 1 · x1
[c(x1)] = 1 · x1
[u(x1)] = 1 + 1 · x1
[f(x1)] = 1 · x1
[t(x1)] = 1 · x1
all of the following rules can be deleted.
u(u(t(t(x1)))) → d(d(c(c(t(t(x1)))))) (30)

1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[n(x1)] =
0
0
+
1 0
0 0
· x1
[d(x1)] =
1
0
+
1 0
0 0
· x1
[o(x1)] =
0
0
+
1 0
0 0
· x1
[c(x1)] =
0
0
+
1 0
0 0
· x1
[u(x1)] =
0
0
+
1 0
0 0
· x1
[f(x1)] =
0
0
+
2 0
0 0
· x1
[t(x1)] =
0
1
+
1 0
0 0
· x1
all of the following rules can be deleted.
f(f(d(d(x1)))) → d(d(f(f(x1)))) (28)

1.1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[n(x1)] =
0
0
+
1 0
0 0
· x1
[d(x1)] =
0
0
+
1 0
0 0
· x1
[o(x1)] =
0
0
+
2 0
0 0
· x1
[c(x1)] =
0
0
+
1 0
0 0
· x1
[u(x1)] =
0
0
+
2 0
0 0
· x1
[f(x1)] =
1
0
+
1 0
0 0
· x1
[t(x1)] =
0
0
+
2 2
2 2
· x1
all of the following rules can be deleted.
u(u(f(f(x1)))) → f(f(u(u(x1)))) (25)

1.1.1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[n(x1)] =
0
0
+
1 1
0 2
· x1
[d(x1)] =
1
1
+
1 0
2 2
· x1
[o(x1)] =
0
0
+
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
+
1 0
0 0
· x1
[t(x1)] =
0
0
+
1 0
0 0
· x1
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.