Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/secr8)

The rewrite relation of the following TRS is considered.

b(b(b(x1))) → a(x1) (1)
a(a(x1)) → a(b(a(x1))) (2)
b(c(x1)) → c(a(a(x1))) (3)
a(c(x1)) → c(b(b(x1))) (4)
a(a(a(x1))) → b(a(a(x1))) (5)

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
b(b(b(x1))) → a(x1) (1)
a(a(x1)) → a(b(a(x1))) (2)
c(b(x1)) → a(a(c(x1))) (6)
c(a(x1)) → b(b(c(x1))) (7)
a(a(a(x1))) → a(a(b(x1))) (8)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(b(x1)) → c#(x1) (9)
c#(b(x1)) → a#(c(x1)) (10)
c#(b(x1)) → a#(a(c(x1))) (11)
c#(a(x1)) → c#(x1) (12)
c#(a(x1)) → b#(c(x1)) (13)
c#(a(x1)) → b#(b(c(x1))) (14)
b#(b(b(x1))) → a#(x1) (15)
a#(a(x1)) → b#(a(x1)) (16)
a#(a(x1)) → a#(b(a(x1))) (17)
a#(a(a(x1))) → b#(x1) (18)
a#(a(a(x1))) → a#(b(x1)) (19)
a#(a(a(x1))) → a#(a(b(x1))) (20)

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

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.