Certification Problem

Input (TPDB SRS_Standard/Zantema_06/11)

The rewrite relation of the following TRS is considered.

a(l(x1)) → l(a(x1)) (1)
a(c(x1)) → c(a(x1)) (2)
c(a(r(x1))) → r(a(x1)) (3)
l(r(a(x1))) → a(l(c(c(r(x1))))) (4)

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
l(a(x1)) → a(l(x1)) (5)
c(a(x1)) → a(c(x1)) (6)
r(a(c(x1))) → a(r(x1)) (7)
a(r(l(x1))) → r(c(c(l(a(x1))))) (8)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
l#(a(x1)) → l#(x1) (9)
l#(a(x1)) → a#(l(x1)) (10)
c#(a(x1)) → c#(x1) (11)
c#(a(x1)) → a#(c(x1)) (12)
a#(r(l(x1))) → l#(a(x1)) (13)
a#(r(l(x1))) → c#(l(a(x1))) (14)
a#(r(l(x1))) → c#(c(l(a(x1)))) (15)
a#(r(l(x1))) → a#(x1) (16)
a#(r(l(x1))) → r#(c(c(l(a(x1))))) (17)
r#(a(c(x1))) → a#(r(x1)) (18)
r#(a(c(x1))) → r#(x1) (19)

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
[l(x1)] = x1 +
2
[c(x1)] = x1 +
0
[a(x1)] = x1 +
2
[r(x1)] = x1 +
2
[l#(x1)] = x1 +
3
[c#(x1)] = x1 +
1
[a#(x1)] = x1 +
2
[r#(x1)] = x1 +
2
together with the usable rules
l(a(x1)) → a(l(x1)) (5)
c(a(x1)) → a(c(x1)) (6)
r(a(c(x1))) → a(r(x1)) (7)
a(r(l(x1))) → r(c(c(l(a(x1))))) (8)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
l#(a(x1)) → l#(x1) (9)
l#(a(x1)) → a#(l(x1)) (10)
c#(a(x1)) → c#(x1) (11)
c#(a(x1)) → a#(c(x1)) (12)
a#(r(l(x1))) → l#(a(x1)) (13)
a#(r(l(x1))) → c#(l(a(x1))) (14)
a#(r(l(x1))) → c#(c(l(a(x1)))) (15)
a#(r(l(x1))) → a#(x1) (16)
r#(a(c(x1))) → r#(x1) (19)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.