Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/matchbox2)

The rewrite relation of the following TRS is considered.

t(o(x1)) → m(a(x1)) (1)
t(e(x1)) → n(s(x1)) (2)
a(l(x1)) → a(t(x1)) (3)
o(m(a(x1))) → t(e(n(x1))) (4)
s(a(x1)) → l(a(t(o(m(a(t(e(x1)))))))) (5)
n(s(x1)) → a(l(a(t(x1)))) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
o(t(x1)) → a(m(x1)) (7)
e(t(x1)) → s(n(x1)) (8)
l(a(x1)) → t(a(x1)) (9)
a(m(o(x1))) → n(e(t(x1))) (10)
a(s(x1)) → e(t(a(m(o(t(a(l(x1)))))))) (11)
s(n(x1)) → t(a(l(a(x1)))) (12)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[e(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[t(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[m(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[o(x1)] =
1 1 0
0 0 0
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[l(x1)] =
1 0 0
0 0 0
1 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[n(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 1 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
a(s(x1)) → e(t(a(m(o(t(a(l(x1)))))))) (11)

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(l) = 5 weight(l) = 1
prec(n) = 4 weight(n) = 3
prec(s) = 2 weight(s) = 1
prec(e) = 3 weight(e) = 3
prec(m) = 1 weight(m) = 3
prec(a) = 6 weight(a) = 1
prec(t) = 0 weight(t) = 1
prec(o) = 7 weight(o) = 3
all of the following rules can be deleted.
o(t(x1)) → a(m(x1)) (7)
e(t(x1)) → s(n(x1)) (8)
l(a(x1)) → t(a(x1)) (9)
a(m(o(x1))) → n(e(t(x1))) (10)
s(n(x1)) → t(a(l(a(x1)))) (12)

1.1.1.1 R is empty

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