Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/x03)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c(x1)] = x1 +
1/3
[b(x1)] = x1 +
1/3
[a(x1)] = x1 +
1
all of the following rules can be deleted.
a(b(c(x1))) → b(x1) (1)

1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(b(c(x1))) → a(x1) (5)
c(x1) → b(x1) (3)
a(a(x1)) → c(a(b(c(x1)))) (6)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(x1) → b#(x1) (7)
b#(b(c(x1))) → a#(x1) (8)
a#(a(x1)) → c#(x1) (9)
a#(a(x1)) → c#(a(b(c(x1)))) (10)
a#(a(x1)) → b#(c(x1)) (11)
a#(a(x1)) → a#(b(c(x1))) (12)

1.1.1.1 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 +
1/3
[b(x1)] = x1 +
1/3
[a(x1)] = x1 +
1
[c#(x1)] = x1 +
0
[b#(x1)] = x1 +
0
[a#(x1)] = x1 +
2/3
together with the usable rules
b(b(c(x1))) → a(x1) (5)
c(x1) → b(x1) (3)
a(a(x1)) → c(a(b(c(x1)))) (6)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(a(x1)) → c#(x1) (9)
a#(a(x1)) → b#(c(x1)) (11)
a#(a(x1)) → a#(b(c(x1))) (12)
and no rules could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.