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 matchbox @ 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 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
o#(t(x1)) → a#(m(x1)) (13)
e#(t(x1)) → s#(n(x1)) (14)
a#(m(o(x1))) → e#(t(x1)) (15)
a#(s(x1)) → o#(t(a(l(x1)))) (16)
a#(s(x1)) → l#(x1) (17)
a#(s(x1)) → e#(t(a(m(o(t(a(l(x1)))))))) (18)
a#(s(x1)) → a#(m(o(t(a(l(x1)))))) (19)
a#(s(x1)) → a#(l(x1)) (20)
s#(n(x1)) → l#(a(x1)) (21)
s#(n(x1)) → a#(x1) (22)
s#(n(x1)) → a#(l(a(x1))) (23)

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

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
[m(x1)] = x1 +
0
[l(x1)] = x1 +
0
[e(x1)] = x1 +
0
[a(x1)] = x1 +
0
[t(x1)] = x1 +
0
[s(x1)] = x1 +
0
[o#(x1)] = x1 +
1
[l#(x1)] = x1 +
0
[e#(x1)] = x1 +
1
[a#(x1)] = x1 +
1
[s#(x1)] = x1 +
1
together with the usable rules
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)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(s(x1)) → l#(x1) (17)
s#(n(x1)) → l#(a(x1)) (21)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.