Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z074)

The rewrite relation of the following TRS is considered.

r(r(x1)) → s(r(x1)) (1)
r(s(x1)) → s(r(x1)) (2)
r(n(x1)) → s(r(x1)) (3)
r(b(x1)) → u(s(b(x1))) (4)
r(u(x1)) → u(r(x1)) (5)
s(u(x1)) → u(s(x1)) (6)
n(u(x1)) → u(n(x1)) (7)
t(r(u(x1))) → t(c(r(x1))) (8)
t(s(u(x1))) → t(c(r(x1))) (9)
t(n(u(x1))) → t(c(r(x1))) (10)
c(u(x1)) → u(c(x1)) (11)
c(s(x1)) → s(c(x1)) (12)
c(r(x1)) → r(c(x1)) (13)
c(n(x1)) → n(c(x1)) (14)
c(n(x1)) → n(x1) (15)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[u(x1)] = 0 · x1 + -∞
[s(x1)] = 0 · x1 + -∞
[b(x1)] = 8 · x1 + -∞
[t(x1)] = 0 · x1 + -∞
[r(x1)] = 0 · x1 + -∞
[c(x1)] = 0 · x1 + -∞
[n(x1)] = 15 · x1 + -∞
all of the following rules can be deleted.
r(n(x1)) → s(r(x1)) (3)
t(n(u(x1))) → t(c(r(x1))) (10)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[u(x1)] = 1 · x1 + -∞
[s(x1)] = 0 · x1 + -∞
[b(x1)] = 0 · x1 + -∞
[t(x1)] = 0 · x1 + -∞
[r(x1)] = 1 · x1 + -∞
[c(x1)] = 0 · x1 + -∞
[n(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
r(r(x1)) → s(r(x1)) (1)
t(r(u(x1))) → t(c(r(x1))) (8)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[u(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 1
0 1 1
0 1 0
· x1 +
1 0 0
1 0 0
0 0 0
[t(x1)] =
1 0 1
1 0 1
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[r(x1)] =
1 1 1
0 0 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[n(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
r(b(x1)) → u(s(b(x1))) (4)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(c) = 7 weight(c) = 4
prec(t) = 6 weight(t) = 2
prec(u) = 0 weight(u) = 5
prec(n) = 4 weight(n) = 2
prec(s) = 2 weight(s) = 5
prec(r) = 3 weight(r) = 1
all of the following rules can be deleted.
r(s(x1)) → s(r(x1)) (2)
r(u(x1)) → u(r(x1)) (5)
s(u(x1)) → u(s(x1)) (6)
n(u(x1)) → u(n(x1)) (7)
t(s(u(x1))) → t(c(r(x1))) (9)
c(u(x1)) → u(c(x1)) (11)
c(s(x1)) → s(c(x1)) (12)
c(r(x1)) → r(c(x1)) (13)
c(n(x1)) → n(c(x1)) (14)
c(n(x1)) → n(x1) (15)

1.1.1.1.1 R is empty

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